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MECHANICS 


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INDUSTRIAL  PHYSICS 


MECHANICS 


BY 

L.  RAYMOND  SMITH 

INSTRUCTOR    IN    INDUSTRIAL    PHYSICS    WM.    L.    DICKINSON    HIGH    SCHOOL, 
JERSEY   CITY,   N.   J.      MEMBER   AMERICAN   SOCIETY   OF   MECHANICAL   ENGINEERS 


FIRST  EDITION 
SECOND  IMPRESSION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
NEW  YORK:  370  SEVENTH  AVENUE 

LONDON:  6  &  8  BOUVERIE  ST.,  E.  C.  4 
1922 


COPYRIGHT,  1922,  BY  THE 
MCGRAW-HILL  BOOK  COMPANY,  INC. 


PRINTED  IN   THE    UNITED  STATES    OF   AMERICA 


T  W  15     MAP  "MS     J-HTCS8     TOBK     P  4 


PREFACE 

The  present  trend  in  education  has  created  a  demand  for  a 
series  of  textbooks  in  which  the  material  presented  is  more 
closely  connected  with  the  every-day  life  of  the  student.  This 
volume  is  the  result  of  an  attempt  to  provide  a  textbook  in 
elementary,  practical  mechanics, — a  textbook  suitable  for  use 
in  technical,  industrial,  vocational  and  evening  schools. 

The  chief  difficulty  encountered  in  writing  an  educational 
book  is  the  choice  of  material.  The  material  presented  here 
consists  largely  of  notes  used  in  class  by  the  author.  It  has 
been  tried  out  and  has  proven  successful.  Certain  time" 
honored  topics,  usually  included  in  similar  books,  have  been 
omitted;  there  has  been  no  sacrifice,  however,  of  fundamental 
principles.  The  order  of  presentation  has  been  found  satis- 
factory. It  may  be  changed,  if  desired,  without  impairing 
the  value  of  the  course.  An  attempt  has  been  made  through- 
out to  keep  the  diction  simple  and  understandable. 

Attention  is  directed  to  the  questions  and  problems  at  the 
end  of  each  chapter  or  division.  The  questions  depend  upon 
the  subject-matter  preceding  and  are  invaluable  both  for 
study  and  review.  The  problems  are  not  difficult  and,  it  is 
hoped,  sufficiently  generous  in  number.  Experience  has 
proven  that  problems  are  an  excellent  medium  for  clarifying 
misunderstandings  on  the  part  of  the  student.  Arithmetic 
is  sufficient  for  most  solutions,  although  a  knowledge  of 
algebra  and  trigonometry  will  be  helpful.  A  special  chapter 
dealing  with  elementary  trigonometry  has  been  included. 

The  author  makes  no  particular  claim  to  originality.  He 
has  consulted  various  standard  works  freely  and  acknowledges 
his  indebtedness  to  many  of  them.  Care  has  been  exercised  to 
keep  the  book  free  from  errors.  In  case  errors  have  crept  in, 
the  author  will  be  glad  to  have  his  attention  called  to  them. 

v 


vi  PREFACE 

Criticism  of  scope  and  content  will  also  be  welcomed. 

Certain  chapters  have  been  made  possible  through  the 
hearty  cooperation  of  various  firms  and  manufacturing  plants 
in  furnishing  information,  photographs,  electrotypes,  etc. 
Grateful  acknowledgment  is  made  to  the  following:  Brown 
and  Sharpe  Mfg.  Co.;  Central  Scientific  Co.;  Cole  Motor  Car 
Co.;  Curtiss  Aeroplane  and  Motor  Corp.;  Dodge  Sales  and 
Engineering  Co.;  Edw.  R.  Ladew  Co.;  Fafnir  Bearing  Co.; 
Ford  Motor  Co.;  Goodell-Pratt  Co.;  Goodyear  Rubber  Co.; 
Goulds  Mfg.  Co.;  Hyatt  Roller  Bearing  Co.;  Keuffel  and 
Esser  Co.;  L.  E.  Knott  and  Co.;  Link-Belt  Co.;  L.  S.  Starrett 
Co.;  Skinner  Engine  Co.;  Timken  Roller  Bearing  Co.;  Whitney 
Mfg.  Co.;  Worthington  Pump  and  Machinery  Corp.;  Yale 
and  Towne  Mfg.  Co. 

The  author  wishes  to  thank  his  colleagues,  T.  Gilbert 
McFadden,  John  H.  Finn  and  Albert  E.  Dickie,  for  the 
valuable  help  they  have  given  in  the  preparation  of  the  manu- 
script. He  also  wishes  to  thank  Frank  E.  Mathewson,  Direc- 
tor of  the  Technical  and  Industrial  Department,  Wm.  L. 
Dickinson  High  School,  for  suggestions  and  advice  incident  to 
the  publication  of  the  book. 

L.  R.  SMITH. 
JERSEY  CITY,  N.  J., 
June,  1922. 


CONTENTS 

PAGE 

PREFACE v 

CHAPTER  I 

INTRODUCTORY  1-6 

Definition  and  scope  of  physics — Definition  and  divisions  of 
mechanics — Matter  and  energy  denned — Constitution  of  matter 
— Kinetic  theory  of  matter — Physical  and  chemical  changes — 
Definition  of  force  and  equilibrium — General  properties  of 
matter — Specific  properties  of  matter. 

CHAPTER  II 

MEASUREMENT  AND  MEASURING  INSTRUMENTS 7-19 

English  and  metric  systems  of  measurement — Metric  tables — 
Common  English-metric  equivalents — The  steel  rule — Adjust- 
able and  non-adjustable  protractors — Ordinary  calipers — The 
speed  indicator — The  slide  rule — The  micrometer  caliper — The 
vernier  caliper. 

CHAPTER  III 

ELEMENTARY  TRIGONOMETRY 20-22 

Trigonometry  defined  and  discussed — Trigonometric  functions 
— Illustrative  trigonometric  solutions. 

CHAPTER  IV 

GRAVITATION  AND  GRAVITY 23-29 

Universal  gravitation — Law  of  universal  gravitation — Gravity 
— Weight — Center  of  gravity — Stable,  unstable  and  neutral 
equilibrium. 

CHAPTER  V 

FORCES 30-34 

Effect  of  forces — Definition  of  a  force — Action  and  reaction — 
Measurement  of  a  force — Graphical  representation  of  forces — 
Moment  of  a  force. 

vii 


viii  CONTENTS 

CHAPTER  VI 

PAGE 

MOTION 35-43 

Motion  defined — Absolute  and  relative  motion — Translatory 
and  rotary  motion — Accelerated  motion — Speed  and  velocity — 
Momentum — Newton's  laws  of  motion — Graphical  representa- 
tion of  motion — Velocity  of  rotation — The  radian. 

CHAPTER  VII 

COMPOSITION  OF  FORCES  AND  VELOCITIES 44-47 

Resultant  of  two  or  more  forces — Parallelogram  of  forces — 
Experimental  verification  of  the  parallelogram  law — Composi- 
tion of  velocities — Directions  for  graphical  work. 

CHAPTER  VIII 

RESOLUTION  OF  FORCES  AND  VELOCITIES 48-51 

Graphical  resolution — Rectangular  components  determined 
graphically — Use  of  squared  paper — Rectangular  components 
determined  trigonometrically. 

CHAPTER  IX       . 

EQUILIBRIUM  OF  CONCURRENT  FORCES 52-56 

Laws  of  concurrent  forces  in  equilibrium — Concurrent  forces 
applied  at  the  same  point — Triangle  of  forces — Concurrent 
forces  applied  at  separate  points. 

CHAPTER  X 

EQUILIBRIUM  OF  PARALLEL  FORCES 57-61 

Tendencies  of  parallel  forces — Laws  of  parallel  forces — Experi- 
mental verification  of  the  laws  of  parallel  forces — Resultant 
and  point  of  application  for  a  system  of  parallel  forces — The 
couple. 

CHAPTER  XI 

EQUILIBRIUM  OF  NON-CONCURRENT  FORCES 62-65 

Conditions  for  equilibrium  of  non-concurrent  forces — The  ladder 
— The  wall  crane. 


CONTENTS  ix 

CHAPTER  XII 

PAGE 

COMMERCIAL  AND  LABORATORY  STRUCTURES 66-76 

General  discussion  of  trusses — Types  of  roof  trusses — Types  of 
bridge  trusses — Laboratory  study  of  a  roof  truss — Laboratory 
study  of  a  stick  and  tie — Laboratory  study  of  a  hoisting  crane 
— Laboratory  study  of  shear  legs. 

CHAPTER  XIII 

ELASTICITY 77-86 

Discussion  and  definition  of  elasticity — Hooke's  law — The 
elastic  limit — The  yield  point — Stress — Strain — Young's 
modulus — Ultimate  strength — Factor  of  safety — Elastic  fatigue 
— Shear  explained — Shearing  stress — Shearing  strain — Modulus 
of  rigidity — Practical  illustrations  of  shear. 

CHAPTER  XIV 

WORK 87-91 

Definition  and  explanation  of  work — Measurement  of  work — 
Units  of  work — Time  and  work — Work  diagrams. 

CHAPTER  XV 

POWER 92-97 

Power  defined  and  discussed — Units  of  power — Brake  horse- 
power— S.A.E.  horsepower — Indicated  horsepower — Mechan- 
ical efficiency. 

CHAPTER  XVI 

ENERGY 98-102 

Nature  of  energy — Fixed  energy — Kinetic  energy — Why  a  body 
possesses  energy — Transformation  of  energy — Conservation  of 
energy — The  mechanical  equivalent  of  heat — The  British 
thermal  unit 

CHAPTER  XVII 

FRICTION 103-111 

Cause  of  friction — Advantages  and  disadvantages  of  friction 
— Coefficient  of  friction — Laws  of  friction — Plain  bearings — 
Ball  bearings — Roller  bearings — Lubrication  of  bearings. 


X  CONTENTS 

CHAPTER  XVIII 

PAGE 

SIMPLE  MACHINES. 112-122 

Definition  and  discussion  of  a  machine — Classification  of 
machines — Input — Output — Efficiency — Velocity  ratio — Mech- 
anical advantage — Law  of  frictionless  machines — Law  of 
non-frictionless  machines — First,  second  and  third  class  levers — 
Pulleys — The  wheel  and  axle — The  inclined  plane — The  screw — 
The  wedge. 

CHAPTER  XIX 

PRACTICAL  STUDY  OF  MACHINES 123-139 

Types  of  chain  blocks — Laboratory  study  of  a  differential  chain 
block — Laboratory  study  of  a  screw-geared  chain  block — 
The  spur-geared  chain  block — Laboratory  study  of  a  chain-drive 
bicycle — Laboratory  study  of  a  jack  screw — The  transmission 
of  an  automobile. 

CHAPTER  XX 

MECHANICAL  TRANSMISSION  OF  POWER 140-160 

Methods  of  power  transmission — Shafts — Couplings — Universal 
couplings  —  Clutches — Cams  — Links — Screw  threads — Chains 
and  sprockets — Pulleys  and  belts — Pulleys  and  ropes — Friction 
wheels — Toothed  gears — Spur  gears — Bevel  gears — Worm 
gears — Helical  gears — Spiral  gears. 

CHAPTER  XXI 

FLUIDS  161-201 

The  three  forms  of  matter — Density — Specific  gravity — Pres- 
sure denned — The  laws  of  liquid  pressure — Transmission  of 
pressure  by  liquids — Pascal's  law  stated  and  illustrated — The 
hydraulic  press — Communicating  columns — Pressure  and  weight 
distinguished — Total  liquid  pressure  on  plane  surfaces — Center 
of  pressure — Dams  and  retaining  walls — Waterwheels — 
Buoyant  force  of  liquids — Archimedes'  principle — The  hydro- 
meter— Characteristics  of  gases — The  atmosphere — Proof  that 
air  has  weight — Proof  that  air  exerts  pressure — Torricelli's 
experiment — Pascal's  experiment — The  mercury  barometer — 
The  aneroid  barometer — The  barograph — Standard  conditions 
of  pressure  and  temperature — The  Magdeburg  hemispheres — 


CONTENTS  xi 

PAGE 

Boyle's  law  stated  and  illustrated — The  aeroplane — The  balloon 
— The  siphon — The  air  pump — The  air  brake — Pumps  for 
liquids — The  pressure  gauge — The  vacuum  gauge — 
Manometers. 

CHAPTER  XXII 

FALLING  BODIES;  CENTRIFUGAL  FORCE;  THE  PENDULUM  .  .  .  202-214 
The  effect  of  gravity  on  falling  bodies — The  acceleration  of 
gravity — Freely  falling  bodies — Bodies  rolling  down  an  incline 
— Bodies  projected  vertically  downward — Bodies  projected  ver- 
tically upward — Bodies  projected  horizontally — Bodies  pro- 
jected at  an  angle  of  elevation  of  less  than  90° — The  nature  of 
centrifugal  force — Effects  and  uses  of  centrifugal  force — The 
formula  for  centrifugal  force — The  pendulum  defined  and  dis- 
cussed— Why  a  pendulum  vibrates — Laws  of  the  pendulum. 

APPENDIX 215-218 

Useful  information — English  and  metric  equivalents — 
Density  of  common  substances — Units  frequently  used — 
Specific  gravity  of  common  substances — Tensile  strength, 
compressive  strength  and  shearing  strength  for  cast  iron, 
wrought  iron  and  mild  steel — Modulus  of  elasticity  (tension, 
compression  and  shear)  for  cast  iron,  wrought  iron  and  mild 
steel — Trigonometric  tables — Decimal  equivalents  of  parts 
of  an  inch. 


INDUSTRIAL  PHYSICS 


MECHANICS 

CHAPTER  I 
INTRODUCTORY 

1.  Physics. — Physics   is   often   defined   as   the   science   of 
matter  and  energy, — the  science  which  attempts  to  explain  the 
relation  between  physical  phenomena  and  the  causes  producing 
them.     Physics    is    conveniently    divided    into    five    parts: 
(1)  mechanics;  (2)  heat;  (3)  light;  (4)  sound;  (5)  electricity. 
This  volume  deals  with  mechanics  only,  a  knowledge  of  which 
is  very  useful  in  stuyding  the  remaining  divisions  of  the 
subject. 

2.  Industrial  Physics. — Industrial  physics  differs  from  the 
so-called  academic  physics  in  that  its  subject  matter  is  more 
practical.     It  substitutes  material  of  live  value  for  certain 
time-honored  topics  which  seem  to  have  little  or  no  connection 
with  the  life  of  the  pupil.     It  is  intended  for  serious,  energetic 
students  who  wish  to  get  a  better  training  than  has  been 
possible  with  the  material  customarily  presented. 

3.  Mechanics. — Mechanics  is  that  branch  of  physical  science 
which  deals  with  forces  and  their  effects.    It  includes  a  study 
of  the  action  of  forces  on  solids,  liquids  and  gases.     Mechanics 
may  be  divided  into  two  distinct  parst:  statics  and  kinetics. 

Statics  deals  with  bodies  so  acted  upon  by  forces  that  no 
motion  results. 

Kinetics  deals  with  bodies  so  acted  upon  by  forces  that 
motion  results. 

l 


2  MECHANICS 

4.  Matter. — Matter  is  anything  which  has  weight  or  which 
takes  up  room.     Matter  may  exist  in  three  different  forms: 
solids,  liquids  or  gases.     Any  material  substance,  such  as  steel, 
gold,  water,  air,  etc.,  is  composed  of  matter. 

5.  Energy. — Energy  is   the   ability   to   do  work.     Gasoline 
possesses  energy  because  it  enables  an  internal  combustion 
engine  to  do  work. 

6.  The   Molecule. — All   matter  is   thought  to   consist   of 
small  particles  called  molecules.    A  molecule  is  defined  as 
the  smallest  particle  into  which  a  substance  may  be  divided 
without  destroying  its  identity.     For  example,  a  molecule  of 
water  is  written  H20;  two  molecules  of  water  is  written 
2H2O,    etc.     Suppose   we   have   one   hundred   molecules   of 
water  or  100  H20;  we  can  then  divide  it  into  100  parts,  each 
part  being  H2O  and  having  the  characteristics  of  water.     If 
we  go  beyond  this  in  our  division,  we  break  up  the  molecule 
into  oxygen  and  hydrogen,  two  gases  which  in  no  way  resemble 
the  original  water. 

Molecules  are  so  minute  that  they  are  not  discernible  with 
the  most  powerful  microscopes.  It  has  been  estimated  by 
Lord  Kelvin1  that,  a  qifuantity  of  water  the  size  of  a  foot- 
ball were  magnified  to  the  size  of  the  earth,  the  molecules 
would  be  between  small  shot  and  footballs  in  size. 

7.  The   Atom. — A   molecule   is   maee   up   of   atoms.     By 
breaking  up  the  water  molecule   (H20)   we  get  the  atoms 
H,  H  and  O.     These  atoms  in  no  way  resemble  the  molecule 
of  which  they  were  formerly  a  part.     The  atom  then,  after 
the  molecule,  is  the  next  step  in  the  division  of  matter  and  is 
the  smallest  particle  of  matter  capable  of  entering  into  combination. 

8.  The  Electron. — According  to  the  most  recent  investiga- 
tions, the  atom  is  a  very  complex  structure.     It  is  thought  to 
be  composed  of  a  nuclues  or  positively  charged  vibrating 
particle  surrounded  by  various  negatively  charged  particles. 

*Sir  William  Thomson  (1824-1907).  One  of  the  greatest  of  the 
nineteenth-century  physicists.  Born  in  Ireland.  Fifty-three  years 
professor  of  physics  at  Glasgow  University,  Noted  for  his  work  in  heat 
and  electricity. 


INTRODUCTORY  3 

These  negative  particles,  called  electrons,  are  very  mobile  and 
possess  tremendous  velocities,  some  approaching  the  velocity 
of  light  (186,000  mi. /sec.)-  So  far  as  is  known,  the  electron  is 
the  smallest  division  of  matter. 

9.  Kinetic  Theory  of  Matter. — There  is  a  well-substantiated 
theory  that  all  molecules  are  in  constant  vibration.     This  is 
known  as  the  kinetic  theory  of  matter. 

10.  Physical  Change. — A  physical  change  is  one  in  which  the 
substance  does  not  lose  its  identity,  such  as  planing  a  piece  of 
steel;  melting  a  quantity  of  tin;  forming  steam  from  water, 
etc.     It  is  evident,  in  each  case,  that  the  new  form  assumed 
has  the  same  molecular  constitution  as  the  original. 

11.  Chemical  Change. — A  chemical  change  is  one  in  which 
the  substance  loses  its  identity.     If  we  burn  a  piece  of  coal,  we 
find  that  it  has  been  changed  into  ash  and  various  gases,  none 
of  which  resemble  the  original  coal.     Another  example  is  the 
rusting  of  a  nail.     Rust  is  a  combination  of  oxygen  and  iron 
and  does  not  resemble  the  iron  in  any  way. 

12.  Force. — A  force  is  anything  that  will  cause  a  body  to 
undergo  a  change  oj  shape  or  a  change  in  condition  oj  motion. 
For  example,  a  force  may  elongate,  compress,  twist  or  bend 
a  body;  it  may  cause  a  body  at  rest  to  move;  it  may  cause  a 
moving  body  to  stop;  it  may  increase  or  decrease  the  velocity 
of  a  body;  or  it  may  cause  a  change  in  the  direction  of  the 
motion. 

13.  Equilibrium. — A  body  at  rest  is  in  equilibrium,  since  its 
state  of  motion  (zero)  is  constant;  a  flywheel  rotating  at  a 
constant  speed  is  in  equilibrium,  since  its  state  of  motion  does 
not  vary.     Equilibrium  means  a  balanced  condition,  in  which 
the  state  of  motion  does  not  change. 

14.  General    Properties    of    Matter. — There    are    certain 
characteristics  that  are  common  to  all  bodies.     Such  char- 
acteristics are  called  general  properties.     A  few  of  the  more 
important  ones  will  be  discussed. 

Porosity. — All  matter  is  porous,  that  is,  a  space  exists 
between  the  molecules  of  all  bodies.  No  two  adjacent  mole- 
cules are  ever  in  contact.  It  has  been  proved  that  gold  is 


4  MECHANICS 

porous  by  forcing  water  through  it  under  heavy  pressure. 
A  teaspoonful  of  sugar  may  be  put  into  a  cup  of  coffee  without 
appreciably  increasing  the  cubical  contents  of  the  cup.  This 
is  explained  by  the  fact  that  the  sugar  molecules  penetrate 
the  inter-molecular  spaces  of  the  coffee.  A  simple  experiment 
will  demonstrate  that  water  is  porous.  A  long  glass  graduate 
is  nearly  filled  with  water  and  the  graduation  coinciding  with 
the  water  line  is  noted.  If  a  few  drops  of  alcohol  are  intro- 
duced into  the  water  and  the  mixture  is  thoroughly  shaken, 
it  will  be  seen  that  the  water  line  has  not  risen.  This  is 
because  the  alcohol  molecules  have  arranged  themselves  in 
the  inter-molecular  spaces  of  the  water. 

Compressibility. — All  bodies  are  compressible.  Liquids  are 
the  least  compressible  and  gases  are  the  most  compressible. 
Solids  vary  widely  in  this  respect.  Steel  is  difficult  to  com- 
press, while  wood  is  relatively  easy  to  compress.  When  a 
body  is  compressed,  the  molecules  are  crowded  closer  together, 
causing  the  body  to  weigh  more  per  unit  volume.  It  is  evident 
that  compressibility  is  a  direct  consequence  of  porosity. 

Indestructibility. — All  matter  is  indestructible.  In  burning 
a  piece  of  coal,  we  do  not  destroy  the  matter  of  which  the  coal 
is  composed.  The  coal  is  converted  into  ash  and  various  gases, 
but  the  original  matter  still  exists  in  other  forms. 

Inertia. — Inertia,  commonly  treated  as  a  force,  is  a  property 
of  matter  causing  a  body  to  resist  any  attempt  to  change  its  condi- 
tion of  rest  or  motion.  If  a  body  is  at  rest,  it  resists  any 
attempt  to  set  it  in  motion.  If  a  body  is  in  motion,  it  resists 
any  attempt  to  stop,  increase  or  decrease  its  speed  or  to  change 
the  direction  of  its  motion. 

Elasticity. — Whenever  a  body  suffers  a  change  of  size  or 
shape,  it  manifests  a  tendency  to  resume  its  original  size 
or  shape.  This  property  is  known  as  elasticity.  Steel  and 
copper  are  highly  elastic,  while  substances  like  rubber  are 
less  elastic.  Even  a  mass  of  putty  is  slightly  elastic.  The 
elasticity  of  a  body  is  measured  by  the  amount  of  its  resistance 
to  a  change  of  size  or  shape. 

Cohesion. — Cohesion  enables  like  molecules  to  hold  together. 


INTRODUCTORY  5 

It  is  greatest  in  solids,  least  in  gases,  and  decreases  with  a  rise 
in  temperature.  Tensile  strength  depends  upon  cohesion  and 
is  measured  by  the  effort  necessarj'  to  rupture  tensionally  a 
body  of  unit  cross-section.  Strictly  speaking,  cohesion  is  a 
property  of  matter.  We  may  think  of  it,  however,  as  the 
force  binding  like  molecules  together. 

Adhesion. — Adhesion  enables  unlike  molecules  to  hold  together. 
Water  clinging  to  a  glass  rod  and  paint  clinging  to  a  piece  of 
wood  are  examples.  Although  adhesion  is  a  property  of 
matter,  we  may  think  of  it  as  the  force  binding  unlike  mole- 
cules together. 

Mass. — Mass  refers  to  the  amount  of  matter  a  body  contains. 
It  is  measured  by  the  resistance  that  a  body  offers  to  a  change 
of  state  of  motion.  The  inertia  of  a  body  is  proportional  to  its 
mass.  Mass  should  not  be  confused  with  weight.  The  weight 
of  a  body  may  vary,  but  the  mass  will  remain  constant.  A 
body  weighing  10  Ib.  at  the  earth's  surface,  will  weigh  practi- 
cally nothing  at  the  earth's  center.  The  amount  of  matter 
will  be  the  same  in  each  case. 

15.  Specific  Properties  of  Matter. — Specific  properties  are 
common  to  certain  bodies  only.  They  enable  us  to  distinguish 
one  substance  from  another.  Following  is  a  discussion  of  a 
few  of  the  more  important  ones. 

Tenacity. — Tenacity  is  the  relative  measure  of  cohesion  in 
various  bodies.  The  tenacity  of  steel  is  very  great;  copper 
is  less  tenacious.  A  carbon  steel  casting  may  require  as  high 
as  80,000  lb./in.2  to  rupture  it  tensionally.  Alloy  steel  is 
even  more  tenacious.  Rolled  copper  has  a  tensile  strength  of 
approximately  40,000  lb./in.2 

Hardness. — Hardness  is  the  resistance  a  body  offers  to  being 
scratched  or  worn  by  another.  A  body  that  will  scratch  another 
is  said  to  be  the  harder  of  the  two.  The  diamond  is  the 
hardest  of  all  known  substances.  If  a  body  is  moving  rapidly, 
it  will  cut  into  another  body  which  is  harder.  The  catting 
ability  of  an  emery  wheel  depends  upon  the  fact  that  the 
emery  is  very  hard  and  that  the  particles  of  emery  are 
moving  with  great  speed. 


6  MECHANICS 

Hardness  should  not  be  mistaken  for  brittleness.  Brittle- 
ness  is  aptness  to  break  under  a  blow.  Vanadium  steel  is 
hard  and  tough,  while  glass  is  hard  and  brittle.  Steel  is  made 
very  hard  and  brittle  by  heating  it  to  a  red  heat  and  then 
plunging  it  into  a  bath  of  water  or  oil.  By  reheating  and 
cooling  slowly,  the  steel  will  become  soft  and  flexible.  The 
latter  process  is  called  annealing.  The  desired  hardness  for 
cutting  tools  is  brought  about  by  tempering.  The  steel  is 
reheated  slowly  until  it  assumes  a  certain  color  and  is  then 
plunged  into  the  bath  of  oil  or  water.  Lathe  tools  are  reheated 
to  a  pale  yellow  color  (approx.  400°F.),  while  saws  are  re- 
heated to  a  dark  blue  color  (approx.  600°F.). 

Ductility. — Ductility  enables  substances  to  be  drawn  out  in 
the  form  of  a  wire.  Platinum  may  be  drawn  into  a  wire 
.00003  of  an  inch  in  diameter.  Glass  is  so  ductile  that  it  may 
be  drawn  out  into  fine  threads  and  the  threads  woven  into 
cloth. 

Malleability. — A  body  which  may  be  hammered,  rolled  or 
pressed  into  thin  sheets  is  said  to  be  malleable.  It  is  possible 
to  make  gold  leaf  .000003  of  an  inch  in  thickness,  in  which 
case  the  gold  is  transparent. 

Questions  and  Problems 

1.  Define  mechanics,  statics,  kinetics,  matter,  energy. 

2.  Distinguish  clearly  between  the  molecule,  atom  and  electron. 

3.  What  is  meant  by  the  kinetic  theory  of  matter? 

4.  Define  and  give  examples  of  a  physical  change. 
6.  Define  and  give  examples  of  a  chemical  change. 

6.  What  is  a  force?     What  effects  may  it  produce? 

7.  What  is  meant  by  equilibrium?     Give  examples. 

8.  What  is  the  difference  between  general  and  specific  properties  of 
matter? 

9.  Define  and  discuss:   porosity,   compressibility,   indestructibility, 
inertia,  elasticity,  cohesion,  adhesion,  mass. 

10.  Define    and    discuss:    tenacity,    hardness,    brittleness,    ductility, 
malleability. 

11.  What  is  meant  by  annealing?     Tempering? 


CHAPTER  II 
MEASUREMENT    AND    MEASURING     INSTRUMENTS 

SECTION  I.     MEASURES  AND  WEIGHTS 

16.  The  English  System. — In  the  United  States  we  still 
employ  the  so-called  English  system  of  measurement  for  every- 
day work.     The  yard  is  taken  as  the  standard  of  length,  the 
pound  as  the  standard  of  mass  (weight)  and  the  second  as  the 
standard  of  time.     In  applied  physics  the  foot  (J<j  of  a  yard). 
the  pound  and  the  second  are  often  used  as  fundamental 
units.     A  system  of  measurement  employing  the  three  latter 
units  is  called  the  foot-pound-second  system.     It  is  usually 
spoken  of  as  the  f.p.s.  system.     The  English  system  is  incon- 
venient and  has   given  way  almost  entirely  to  the  metric 
system  for  purely  scientific  work.     At  present  there  seems  to 
be  a  tendency  toward  the  adoption  of  the  metric  system  in  this 
country  for   all   purposes.     The   coming   generation   should 
thoroughly  master  the  metric  system,  so  that  there  will  be  as 
little  confusion  as  possible  when  the  change  occurs. 

17.  The  Metric  System. — About  the  time  of  the  French 
Revolution,  the  government  of  France  appointed  a  commission 
to  devise  a  system  of  weights  and  measures  to  replace  the 
awkward  system  then  in  use.     This  resulted  in  the  establish- 
ment of  the  metric  system  in  France.     It  has  since  been  made 
compulsory  in   most   civilized   countries,   England   and  the 
United    States    being    exceptions.     The    metric    system    is 
extremely  easy  to  use  on  account  of  its  simplicity  and  decimal 
scale.     In  scientific   work  the  centimeter,   gram  and  second 
are  used  as  fundamental  units.     A  system  of  measurement 
employing  the  three  latter  units  is   called  the  centimeter- 
gram-second  system.     It  is  usually  referred  to  as  the  c.g.s. 
system. 

7 


8 


MECHANICS 


The  meter  (m.)  was  adopted  as  the  standard  of  length  in 
the  metric  system.  It  was  intended  to  be  Ko>ooo>ooo  of  the 
distance  from  the  Equator  to  the  North  Pole,  measured  on  the 
meridian  of  Paris.  This  distance  was  computed  incorrectly, 
however,  and  the  standard  meter  to-day  is  the  distance 
between  two  transverse  scratches  on  a  bar  of  platinum- 
iridium  at  a  temperature  of  0°C.  This  bar  is  preserved  in  the 
archives  of  France  and  replicas  have  been  sent  to  all  civilized 
countries.  The  meter  is  39.37  inches  long.  Our  yard  is 
officially  defined  as  3>60M,937  of  a  meter.  The  meter  is 
divided  into  10  equal  parts  or  decimeters;  100  equal  parts  or 
centimeters;  and  1,000  equal  parts  or  millimeters.  The 
centimeter,  due  to  its  more  convenient  size,  is  universally  used 


12345 
Centimeter 


i ' 

Inches 
I  2 

FIG.   1. — Comparison  of  the  centimeter  and  inch. 


in  scientific  work.     Figure  1  shows  the  relation  of  the  centi- 
meter and  inch. 

The  metric  standard  of  mass  (weight)  is  the  kilogram  (Kg.). 
It  was  intended  to  be  the  mass  of  one  cubic  decimeter  (dm.3)  of 
pure  water  at  a  temperature  of  4°C.  Due  to  a  very  slight  error, 
this  relation  is  not  exactly  true.  It  is  close  enough,  however, 
for  all  practical  purposes.  The  prototype  kilogram  is  a 
cylinder  of  platinum-iridium  kept  in  the  archives  of  France, 
copies  of  which  have  been  furnished  to  all  civilized  countries. 
The  gram  (g.)  or  one  thousandth  of  a  kilogram  is  taken  as  the 
fundamental  unit.  The  gram  is  divided  into  10  equal  parts  or 
decigrams;  100  equal  parts  or  centigrams;  and  1,000  equal 
parts  or  milligrams.  Our  pound  avoirdupois  is  equivalent  to 
453.5924277  grams.  For  practical  work,  454  grams  is  used. 
The  kilogram  is  equivalent  to  about  2.2  pounds.  Figure  2 
shows  the  practical  relation  of  the  pound  and  kilogram. 


MEASURING  INSTRUMENTS 

The  unit  of  capacity  in  the  metric  system,  both  dry  and  wet, 
is  the  liter  (1).  It  is  equivalent  to  the  volume  of  a  cube  one 
decimeter  on  a  side  and  is  approximately  the  same  as  a  quart. 
The  liter  is  divided  into  10  equal  parts  or  deciliters;  100  equal 
parts  or  centiliters;  and  1,000  equal  parts  or  milliliters.  For 
practical  purposes,  a  liter  of  water  may  be  considered  to 


FIG.  2. — One   kilogram  weighs  2.2  pounds  and  just  balances   1,000   cubic 
centimeters  of  water. 

weigh  one  kilogram.  Hence  one  cubic  centimeter  of  water 
weighs  one  gram. 

The  unit  of  time  in  the  metric  system  is  the  second.  It  is 
defined  as  3^6,400  of  the  mean  solar  day. 

The  following  tables  will  familiarize  the  student  with  the 
essential  parts  of  the  metric  system.  Heavy  type  is  used  for 
the  more  common  units. 


MEASURES  OF  LENGTH 


10  millimeters  (mm 
10  centimeters 
10  decimeters 
10  meters 
10  dekameters 
10  hectometers 
10  kilometers 


=  1  centimeter  (cm.) 
=  1  decimeter  (dm.) 
=  1  meter  (m.) 
=  1  dekameter  (Dm.) 
=  1  hectometer  (Hm.) 
=  1  kilometer  (Km.) 
=  1  myriameter  (Mm.) 


10  MECHANICS 

MEASURES  OF  SURFACE 

100  square  millimeters  (mm.2)  =  1  square  centimeter  (cm.2) 
100  square  centimeters  =  1  square  decimeter  (dm.2) 

100  square  decimeters  =  1  square  meter  (m.2) 

100  square  meters  =  1  square  dekameter  (Dm.2) 

100  square  dekameters  =  1  square  hectometer  (Hm.2) 

100  square  hectometers  =  1  square  kilometer  (Km.2) 

MEASURES  OF  VOLUME 

1000  cubic  millimeters  (mm.3)  =  1  cubic  centimeter  (cm.3) 
1000  cubic  centimeters  =  1  cubic  decimeter  (dm.3) 

1000  cubic  decimeters  =  1  cubic  meter  (m.3) 

MEASURES  OF  CAPACITY 

10  milliliters  (ml.)  =  1  centiliter  (cl.) 

10  centiliters  =  1  deciliter  (dl.) 

10  deciliters  =  1  liter  (1) 

10  liters  =  1  dekaliter  (Dl.) 

10  dekaliters  =  1  hectoliter  (HI.) 

10  hectoliters  =  1  kiloliter  (Kl.) 

MEASURES  OF  WEIGHT 

10  milligrams  (mg.)  =  1  centigram  (eg.) 

10  centigrams  =  1  decigram  (dg.) 

10  decigrams  =  1  gram  (g.) 

10  grams  =  1  dekagram  (Dg.) 

10  dekagrams  =  1  hectogram  (Hg.) 

10  hectograms  =  1  kilogram  or  kilo  (Kg.) 

10  kilograms  =  1  myriagram  (Mg.) 

10  myriagrams  =  1  quintal  (Q) 

10  quintals  =  1  tonneau  or  metric  ton  (T.) 

The  following  equivalents  should  be  memorized  by  the  pupil. 
They  are  sufficiently  accurate  for  all  ordinary  work. 

1  meter         =  39 . 37  inches  (exact) 

1  inch  =    2.54  centimeters  (approx.) 

1  kilometer  =       .62  mile  (approx.) 

1  kilogram    =    2.2    pounds  (approx.) 

1  liter  =     1.06  liquid  quarts  (approx.) 


MEASURING  INSTRUMENTS  11 

QUESTIONS  AND  PROBLEMS 

1.  Give  a  brief  history  of  the  metric  system. 

2.  Show  why  the  metric  system  should  be  made  compulsory  in  this 
country. 

3.  What  is  the  f.p.s.  system  of  measurement?    The  c.g.s.  system? 

4.  Define:  yard,  foot,  pound,  meter,  kilogram,  gram,  liter  and  second. 
6.  Why  must  the  prototype  meter  be  kept  at  a  temperature  of  0°C.  ? 

6.  What  is  the  weight  in  grams  of  10  cm.3  of  water? 

7.  On  July  2,  1901,  the  Cornell  Varsity  Crew  at  Poughkeepsie  rowed 
4  mi.  in  18  min.,  53K  sec.,  establishing  the  record  for  the  course.     What 
was  the  average  speed  in  meters  per  sec?     In  kilometers  per  hr? 

8.  The  amateur  record  for  the  hundred  yard  dash  is  9^  sec.     Express 
the  speed  in  meters  per  sec.     In  kilometers  per  min. 

9.  The  muzzle  velocity  of  a  projectile  was  2,000  ft.  per  sec.     How 
many  meters  per  hr.  was  this? 

10.  An  aviator  ascended  to  a  height  of  30,000  ft.     State  the  altitude 
in  meters. 

11.  Light  travels  with  a  velocity  of  300,000  kilometers  per  sec.     Give 
the  equivalent  velocity  in  feet  per  sec. 

12.  If  the  barometer  reads  760  millimeters,  what  will  it  read  in  inches? 

13.  The  wheel  base  of  an  automobile  is  120  inches.     What  is  it  in 
centimeters? 

14.  The  diameter  of  the  earth  is  about  8,000  miles.     Express  the 
diameter  and  circumference  of  the  earth  in  kilometers. 

15.  A  fly  wheel  two  meters  in  diameter  rotates  at  the  rate  of  200 
r.p.m.     What  linear  distance  will  a  point  on  the  circumference  travel  in 
4  minutes? 

16.  An  iron  casting  weighs  200  Ib.     What  is  its  weight  in  kilograms? 
in  metric  tons? 

17.  An  aluminum  block  weighs  10  Kg.     What  is  the  weight  in  pounds? 

18.  How  many  cubic  centimeters  are  there  in  a  granite  block  2.5 
meters  on  a  side? 

19.  Compute  the  number  of  cubic  inches  in  a  cubic  meter. 

20.  A  gasoline  tank  holds  15  gallons.     Express  its  capacity  in  liters. 

21.  A  tank  is  2  X  2  X  6  meters.     How  many  grams  of  water  will  it 
hold? 

22.  The  drip  vat  of  a  steam  power  plant  is  2  X  2  X  3  meters.     How 
many  cubic  feet  of  drip  will  it  hold? 

23.  How  many  quarts  are  there  in  a  cubic  meter? 

SECTION  II.     MEASURING  INSTRUMENTS 

18.  The  Steel  Rule.— A  6  in.  flexible  steel  rule,  graduated  in 
sixty-fourths  of  an  inch  on  one  side  and  millimeters  on  the 


12 


MECHANICS 


other,  should  be  the  personal  property  of  each  student.  It 
is  not  expensive  to  buy,  is  practically  indestructible,  and  may 
be  carried  in  the  pocket. 


FIG.  3.- — Steel  rule. 


19.  The  Protractor.— Protractors  are  used  to  measure  angles 
and  are  either  adjustable  or  non-adjustable.  One  or  more 
large  adjustable  protractors  should  be  in  every  laboratory, 
and  the  student  should  own  a  small  brass  or  steel  protractor 


FIG.  4. — Ordinary  form  of  protractor. 


FIG.  5. — Adjustable  protractor. 

of  the  non-adjustable  type.  The  exact  method  of  measuring 
an  angle  may  be  seen  from  a  study  of  the  accompanying 
diagrams. 

20.  The  Ordinary  Caliper. — Calipers,  shown  herewith,  are 
used  in  taking  external  and  internal  measurements  when  a 
high  degree  of  accuracy  is  not  necessary.  They  are  either 
of  the  "spring"  or  "firm-joint"  type.  Outside  and  inside 


MEASURING  INSTRUMENTS  13 

calipers,  of  different  capacities,  should  always  be  available 
for  laboratory  work. 


Outside  Inside 

FIG.  6. — Spring  calipers. 


Outside  Inside 

FIG.  7. — Firm-joint  calipers. 

21.  The  Speed  Indicator. — The  speed  indicator  is  an  instru- 
ment for  measuring  the  number  of  revolutions  of  shafts  and 
spindles.  The  Starrett  indicator  shown  here  registers  in  either 


14 


MECHANICS 


direction  up  to  100.  The  rotating  disc  carries  a  raised  knob 
which  should  coincide  with  the  knob  on  the  dial  at  the  start. 
The  disc  is  carried  around  by  friction ;  hence  it  is  an  easy  matter 
to  set  it  back  to  zero.  Each  complete  rotation  of  the  disc 
indicates  100  rotations  of  the  shaft  or  spindle  under  test.  In 


FIG.  8. — Speed  indicator. 

principle,  the  indicator  is  a  worm  and  worm  wheel.  The 
student  will  be  able  to  use  the  indicator  without  detailed 
instructions. 

22.  The  Slide  Rule. — The  slide  rule  is  a  device  to  facilitate 
the  solution  of  numerical  problems.  Its  accuracy  is  sufficient 
for  most  kinds  of  work.  Problems  involving  multiplication, 
division,  roots,  powers,  trigonometric  functions,  etc.,  may  be 
solved  with  a  great  saving  of  time.  Students  will  find  the 


FIG.  9. — Polyphase  slide  rule. 

slide  rule  of  great  convenience  for  both  home  work  and  class 
work.  It  is  comparatively  expensive,  yet  it  is  hoped  that  a 
majority  of  students  will  be  able  to  own  one.  The  Polyphase 
rule  shown  in  the  diagram,  has  proven  very  satisfactory. 
Complete  and  detailed  instructions  are  furnished  by  the 
manufacturer  with  each  rule. 

23.  The  Micrometer  Caliper. — The  micrometer  caliper  is 
used  to  take  measurements  when  very  fine  readings  are 
desired.  The  type  shown  in  Fig.  10,  is  for  external  measure- 
ments. Micrometer  calipers  may  be  procured  in  various 


MEASURING  INSTRUMENTS 


15 


sizes,  some  reading  as  high  as  24  in.  Ordinarily,  they  read 
only  to  thousandths  of  an  inch,  but  by  careful  estimation  of 
scale  divisions,  finer  readings  are  obtainable.  They  are  also 
graduated  metrically  with  corresponding  accuracy.  The 


FIG.  10. — Micrometer  caliper. 

one  described  here  will  read  accurately  to  a  thousandth  of  an 
inch. 

The  micrometer  caliper  consists  of  the  following  essential 
parts:  the  frame  (F);  the  anvil  (A);  the  spindle  (£);  the 
sleeve  (H);  and  the  thimble  (T). 

The  sleeve  is  graduated  in  teriths  of  an  inch.  The  gradua- 
tions are  labelled  1,  2,  3,  etc.  Each  numbered  division  is 
sub-divided  into  four  equal  parts  or  fortieths  of  an  inch.  The 
spindle  has  forty  threads  to  the  inch  and  rotates  in  a  fixed  nut. 


Ltuilii 

15 
10                    § 

c                                 >5 

==^ 

(a) 

.  o  ,£.  _  4" 
*  ~To 


p       2 

To  4b  ^fooo  ~  1000 


iooo 

.400+.000+.OOO=.400 

FIG.  11.  —  Reading  a  micrometer  caliper. 

Hence  one  full  rotation  of  the  thimble  will  cause  the  spindle  to 
move  one  fortieth  of  an  inch  to  the  left  or  right.  Since  the 
bevelled  edge  of  the  thimble  is  divided  into  twenty-five  equal 
parts,  a  movement  of  one  scale  division  on  the  bevel  will  cause 
a  spindle  movement  of  one  twenty-fifth  of  one  fortieth  or  one 
thousandth  of  an  inch. 


10 


MECHANICS 


In  using  the  micrometer  caliper,  the  body  to  be  measured 
should  be  placed  in  the  jaws  of  the  instrument  and  the  thimble 
rotated  until  light  contact  is  made.  If  the  caliper  is  forced, 
inaccurate  readings  will  result.  A  ratchet  device,  automatically 
stopping  the  spindle  at  the  proper  pressure,  is  of  ten  provided. 
The  " spindle"  reading  plus  the  ^thimble"  reading  will  give 
the  dimension  sought. 

The  diagrams  in  Fig.  1 1  represent  the  relative  positions  of 
the  spindle  and  thimble  for  the  readings  indicated.  The 
student  should  study  these  diagrams  very  carefully. 

24.  The  Vernier  Caliper. — The  vernier  caliper,  like  the 
micrometer  caliper,  is  used  when  very  fine  measurements  are 
required.  The  one  shown  below  will  read  down  to  one 
c  D 


FIG.  12. — Vernier  caliper. 


thousandth  of  an  inch.  It  is  graduated  to  read  in  thousandths 
of  an  inch  on  the  front  and  sixty-fourths  of  an  inch  on  the  back. 
Only  the  front  will  be  discussed. 

Figure  12  shows  the  general  construction  of  a  standard 
vernier  caliper.  It  consists  of  the  following  parts:  the  beam 
(A) ;  the  slide  (B) ;  the  binding  screws  (C  and  D) ;  the  adjusting 
screw  (E) ;  and  the  jaws  (F  and  G) . 

The  beam  (Fig.  13)  is  graduated  in  inches  and  is  numbered 
1,  2,  3,  etc.,  large  numbers  being  used.  The  inches  are 
divided  into  ten  equal  parts  or  tenths  of  an  inch  and  are  num- 
bered 1,  2,  3,  etc.,  small  numbers  being  used.  The  tenths  are 
further  divided  into  four  equal  parts  or  fortieths  of  an  inch  and 
are  not  numbered.  The  smallest  division  on  the  beam  is, 
therefore,  one  fortieth  (.025)  of  an  inch. 

The  slide  (Fig.  13)  is  divided  into  twenty-five  equal  parts 
and  is  numbered  5,  10,  15,  20  and  25.  Each  numbered  part 


MEASURING  INSTRUMENTS 


17 


is  sub-divided  into  five  equal  parts  without  numbers.     There 
are  in  all,  then,  twenty-five  small  divisions  on  the  slide. 

By  reference  to  Fig.  14,  we  see  that  twenty-four  small 
divisions  on  the  beam  correspond  in  length  to  twenty-five 
small  divisions  on  the  slide.  Hence  a  small  division  on  the 
slide  =  2^5  of  a  small  division  on  the  beam  =  2J^5  of 
.02,5  in.  =  .024  in.  Further,  we  see  that  the  first  graduation  on 


FIG.  13. — Vernier  caliper,  view  of  beam  and  slide. 


Small  Graduations  =  .  O2S ' 


0           1 

234 

5          GE1         J 

1  1  1 

| 

| 

1  1  1 

1  1  1 

| 

ii 

I 

1  1  1 

1      1 

1      1  1  1 

I  I 

in 

( 

3            5 

1 
10            15 

20          2 

s 

SLIDE 

JEAM 


Small  6raduai-ions=  .  0?4  " 
FIG.   14. — Vernier  caliper,   comparison  of  small  graduations  on  beam  and 

slide. 


the  slide  differs  from  the  first  graduation  on  the  beam  by  .001  in.  ; 
that  the  second  graduation  on  the  slide  differs  from  the  second 
graduation  on  the  beam  by  .002  in.,  etc. 

The  following  instructions  should  be  observed  in  manipulat- 
ing the  vernier.  The  instructions  are  intended  only  for 
external  measurements  and  refer  to  Fig.  12.  After  moving 
the  slide  over  until  both  jaws  nearly  touch  the  object,  the 
binding  screw  (Z>)  should  be  screwed  down.  The  fine  adjust- 
ment (E)  should  be  rotated  until  each  jaw  touches  the  object 
2 


18 


MECHANICS 


lightly.     After  screwing  down  the  binding  post  (C)  so  as  to 
lock  the  jaws,  the  reading  may  be  taken. 

To  illustrate  the  reading  of  a  vernier  caliper,  let  us  take  an 
example  and  refer  to  Fig.  15.  In  measuring  the  diameter  of  a 
steel  rod  it  was  found  that  the  twenty-ninth  small  division  on 
the  beam  exactly  lined  up  with  the  eighteenth  small  division 
on  the  slide.  It  is  evident  that  the  diameter  of  the  rod  is  the 
linear  distance  represented  by  the  twenty-nine  small  divisions 
on  the  beam  less  the  linear  distance  represented  by  the  eighteen 
small  divisions  on  the  slide,  viz.,  .725  in.  —  .432  in.  =  .293  in. 

BEAM 


10 


SLIDE 
FIG.  15. — Vernier  caliper,  showing  coincidence  of  graduations  X  and  Xi. 

Since  the  zero  mark  on  the  slide  (Fig.  15)  extends  somewhat 
beyond  the  .275  in.  mark  on  the  beam,  it  will  be  seen  that  the 
diameter  of  the  rod  is  also  .275  in.  plus  the  distance  to  zero  on 
the  slide.  As  the  points  of  coincidence  (X  and  X 0  are  eighteen 
slide  divisions  beyond  zero  on  the  slide,  then  zero  on  the  slide 
must  extend  .018  in.  beyond  the  .275  in.  mark  on  the  beam. 
Hence  the  correct  reading  will  be  .275  in.  +  .018  in.  =  .293  in., 
the  same  as  determined  in  the  first  method.  The  method  last 
described  should  always  be  used  as  it  is  more  direct. 

To  read  the  vernier  caliper  we  may  adopt  the  following 
rule:  Note  and  record  the  first  small  division  on  the  beam  to  the 
left  of  zero  on  the  slide.  Add  to  this  .001  in.  for  each  small 
slide  division  up  to  the  point  where  the  beam  and  slide  divisions 
coincide. 

QUESTIONS  AND  PROBLEMS 

1.  Describe  two  types  of  calipers  used  for  approximate  work. 

2.  For  what  is  the  micrometer  caliper  used?     Describe  its  construc- 
tion and  operation. 


MEASURING  INSTRUMENTS  19 

3.  For  what  is  the  vernier  caliper  used?     Describe  its  construction 
and  operation. 

4.  The  following  readings  were  obtained  with  a  micrometer  caliper: 
.240  in.;  1.361  in.;  and  2.546  in.     What  is  the  sleeve  and  thimble  reading 
in  each  case? 

6.  If  the  above  readings  were  obtained  with  a  vernier  caliper,  state 
in  each  case  the  point  at  which  the  beam  and  slide  division  coincided. 


CHAPTER  III 
ELEMENTARY  TRIGONOMETRY 

25.  Trigonometry  Defined — Trigonometry  deals  with  the 
relations  which  the  sides  and  angles  of  a  right-angled  trianler 
bear  to  each  other.     An  elementary  knowledge  of  trigonometgy 
facilitates  the  solution  of  many  numerical  problems.     Hence  it 
is  very  important  for  the  student  to  gain  a  working  knowled  eg 
of  the  subject. 

26.  Trigonometric  Functions. — A  trigonometric  function  is 

the  ratio  which  one  side  of  a  right-angled 
triangle  bears  to  another.  There  are  six 
such  relations.  They  are  given  here  with 
respect  to  angle  0.  The  first  three 
functions  are  sufficient  for  ordinary 
FIG.  16.  purposes. 

BC       side  opposite  ,  .  ,     . 

-7-n  ~  ~r~  ~  —  sine  (sin)  of  angle  0. 

AB         hypotenuse 

AC       side  adjacent  ,      .     . 

-1-5  =  -v—  -  =  cosine  (cos)  of  angle  6. 

AB        hypotenuse 

BC       side  opposite  .  ,.     .     , 

-779  =  — TJ — ~—    -  =  tangent  (tan)  of  angle  8. 

AC      side  adjacent 

AC       side  adjacent  .  .     .     ,.        , 

^79  =  -TJ 77-  =  cotangent  (cot)  of  angle  0. 

BC      side  opposite 

AB        hypotenuse  .  ,     .    f 

•779  =  -T-] — -T. =  secant  (sec)  of  angle  8. 

AC       side  adjacent 

AB        hypotenuse 

HTF  =  TJ          — =r    =  cosecant  (cosec)  of  angle  8. 

BC       side  opposite 

.'    27.  Trigonometric  Solutions. — Let  us  illustrate  the  solution 
of  problems  by  the  use  of  the  sine,  cosine  and  tangent.     Refer- 

20 


ELEMENTARY  TRIGONOMETRY  21 

ring  to  Fig.  17,  we  have  a  right  angled-triangle.  If  the  side 
J5Cis5  inches  in  length,  by  actual  measurement  the  other  side 
will  be  8.66  inches  and  the  hypotenuse  10  inches.  As  long  as 
the  angles  are  kept  constant,  even  if  the  lengths  of  the  sides 
are  changed,  there  will  always  exist  the  same  numerical  ratio 
between  the  lengths. 

Solutions. — 1   Suppose  it  is  desired  to    find 
BC,  knowing  the  angle  BAG  and  the  hypotenuse 

AB.  Solution:  BC/AB  =  sin  BAG.     BC  =  AB 
sin  BAC.     Referring  to  the  trigonometric  tables 
in  the  appendix,  we  find  that  the  sine  of  angle 
BAG  (30°)  is  .500.     Hence,   by   substitution, 
BC  =  10  X  .500  =  5. 

2.  Suppose  it  is  desired  to  find  AC,  knowing  angle  BAC  and  the 
hypotenuse  AB.     Solution:  AC/AB  =  cos  BAC.     AC  =  AB  cos  BAC. 
By  substitution,  AC  =  10  X  .866  =  8.66. 

3.  Suppose  it  is  desired  to  find  BC,  knowing  angle  BAC  and  the  side 

AC.  Solution:  BC/AC  =  tan  BAC.     BC  =  AC  tan  BAC.     By  substi- 
tution, BC  =  8.66  X  .577  =  4.996.     (Since  the  tables  in  the  appendix 
are  to  three  decimal  places  only,  we  get  4.996  instead  of  5  as  expected. 
Three  place  tables  are  accurate  enough  for  ordinary  purposes.) 

4.  Suppose  it  is  desired  to  find  angle  BAC,  knowing  sides  AB  and  BC. 
Solution:  Sin  BAC  =  BC/AB.    Sin  BAC  =  Ko  =  -500.     By  reference 
to  the  trigonometric  tables,  it  is  found  that  the  angle  to  which  .500 
corresponds  is  30°. 

Questions  and  Problems 

1.  With  what  does  trigonometry  deal?     Why  is  it  important  in 
mechanics? 

2.  What  is  a  trigonometric  function? 

3.  Name  and  define  the  six  trigonometric  ratios. 

4.  A  guy  wire  attached  to  the  top  of  a  telephone  pole  is  secured  to  the 
ground  by  means  of  a  "dead  man"  and  makes  an  angle  of  60°  with  the 
ground.     The  wire  enters  the  ground  18  ft.  from  the  bottom  of  the  pole. 
(a)  How  long  is  the  wire?     (6)  How  high  is  the  pole? 

6.  A  flag  pole  rope,  when  pulled  20  ft.  away  from  the  bottom  of  the 
pole,  just  touches  the  ground.  If  the  rope  pulley  is  40  ft.  above  the 
ground,  (a)  how  long  is  the  rope?  (6)  What  angle  does  it  make  with 
the  ground? 

6.  The  Woolworth  building  is  792  ft.  high.  An  imaginary  line,  at  an 
angle  of  60°  to  the  vertical,  is  drawn  from  the  top  of  the  tower  to  the 
ground.  Assuming  no  change  in  altitude  and  no  curvature  of  the  earth 


22  MECHANICS 

(a)  how  far  horizontally  from  a  point  directly  under  the  tower  would  the 
line  strike  the  ground?     (6)  How  long  would  the  line  be? 

7.  A  ladder  30  ft.  long  is  placed  against  a  house  at  an  angle  of  48° 
to  the  horizontal,     (a)  How  high  is  the  top  of  the  ladder  above  the 
ground?     (6)  How  far  is  the  bottom  of  the  ladder  from  the  house? 

8.  A  shadow  cast  by  the  top  of  a  monument  30  ft.  high  is  12  ft.  long. 
What  is  the  angle  of  the  sun  with  reference  to  the  ground? 

9.  A  captive  balloon  200  meters  high  is  driven  by  the  wind  until  the 
tie  rope  is  at  an  angle  of  80°  to  the  ground.     What  is  the  second  height 
of  the  balloon  ? 

10.  From  a  light  house  50  ft.  high,  two  rocks  are  seen  in  a  direct  line 
with  the  observer.  If  the  rocks  make  angles  of  60°  and  70°  respectively 
with  the  vertical  at  the  top  of  the  light  house,  how  far  apart  are  the  rocks? 


CHAPTER  IV 
GRAVITATION  AND  GRAVITY 

28.  Universal  Gravitation.  —  It  has  been  found  that  there  is 
a  mutual  attraction  existing  between  all  bodies  in  the  universe. 
For  instance,  there  is  an  attraction  between  the  masses  of  the 
earth  and  the  moon,  although  they  are  separated  by  a  distance 
of  over  240,000  miles.  The  periodic  rising  and  falling  of  the 
ocean  waters,  known  as  tides,  is  ascribed 
largely  to  the  gravitational  effect  of  the  C\  d  C\ 


moon.  There  is  also  an  attraction 
between  the  earth  and  the  sun,  the 
earth  and  the  stars,  etc.  The  effect  is 
slight,  however,  due  to  the  tremendous 
distances  between  them. 

29.  The  Law  of  Universal  Gravita-  ?M 

tion.  —  A  study  of  gravitation  led  to  an 
enunciation  by  Sir  Isaac  Newton1  of 
the  following  law  : 

Between  every  two  particles  of  matter 
in  the   universe  there  exists    a  mutual 
attraction  which  varies  directly   as   the  ,£ 
product  of  the  masses  and  inversely  as  middle  diagram  is  four 
the  square  of  the  intervening  distance.       iim™  as  gr?at  as  in  upper 

*  J  y  and  lower  diagrams. 

The  above  law  is  called  the  law  of 

universal  gravitation  and  will  be  readily  understood  by  refer- 
ence to  Fig.  18.  A  and  B  (upper  diagram)  represent  two 
bodies  with  masses  of  M  and  m  respectively  and  separated 

JSir  Isaac  Newton  (1642-1727).  Born  in  England.  Buried  in 
Westminster  Abbey.  Famous  as  a  physicist,  mathematician  and 
astronomer.  Professor  of  mathematics  at  Cambridge,  member  of 
Parliament  and  Master  of  the  Mint.  Responsible  for  the  law  of  gravi- 
tation, the  three  laws  of  motion  which  bear  his  name,  the  bionomial 
theorem,  the  calculus  and  many  discoveries  in  light. 


24  MECHANICS 

by  a  distance  of  d.  The  product  of  the  masses  is  Mm.  If 
the  masses  are  doubled  (middle  diagram),  their  product  will 
be  4  Mm  and  the  mutual  gravitational  force  will  be  increased 
4  times.  If,  in  addition,  the  distance  between  A  and  B  is 
doubled  (lower  diagram),  the  gravitational  attraction  between 
them  will  be  reduced  to  }/±  its  former  value.  The  attraction 
in  the  upper  and  lower  diagrams  is  therefore  identical.  If 
the  distance  is  tripled  and  the  masses  doubled  in  the  upper 
diagram,  the  attractive  force  will  be  ^  of  its  original  intensity. 

30.  Gravity. — The  term  gravity  refers  to  gravitation  in  a 

restricted  sense.  Specifically,  it  means  the  mutual 
attraction  between  the  earth  and  bodies  on  or  near  the 
earth's  surface.  Thus  the  force  which  impels  bodies 
toward  the  earth  is  called  the  force  of  gravity. 

The  force  of  gravity  tends  to  pull  all  objects  towards 
the  earth's  center  of  mass.  This  point  is  approxi- 
mately at  the  geometrical  center  of  the  earth  (4,000 
miles  below  the  surface).  The  direction  of  the  pull  is 
conveniently  determined  by  means  of  a  plumb  line. 
Ordinarily  the  plumb  line  points  directly  toward  the 
earth's  center  of  mass.  It  will  be  deflected  slightly, 
however,  if  brought  near  a  large  mountain.  A  deflec- 
tion will  also  take  place  if  the  cord  is  too  long,  due  to 
FIG.  19.  the  rotation  of  the  earth.  Ordinarily,  deflections  are 
Piin™b  to°  slight  to  have  any  serious  effects  and,  for  practical 
purposes,  we  may  assume  that  the  plumb  line,  if 
extended,  would  pass  through  the  center  of  the  earth.  We 
say  that  the  plumb  line  assumes  a  vertical  position.  Any  line 
drawn  perpendicular  to  the  plumb  line  is  said  to  be  horizontal. 

31.  Weight. — The  mass  or  quantity  of  matter  possessed  by 
a  body  remains  constant,  but  the  weight  is  subject  to  variation. 
The  weight  of  a  body  is  the  mutual  attraction  existing  between 
the  body  and  the  earth.     For  example,  if  a  man  weighs  175  Ib. 
at  a  certain  place  on  the  earth,  the  earth  is  pulling  the  man 
toward  its  center  with  a  force  of  175  Ib.;  likewise,  the  man  is 
exerting  a  pull  of  175  Ib.  upon  the  earth.     The  maximum 
weight  is  at  the  earth's  surface. 


GRAVITATION  AND  GRAVITY  25 

32.  Bodies  Above  and  Below  the  Earth's  Surface. — From 
the  law  of  universal  gravitation,  it  follows  that  the  weight  of 
a  body  above  the  earth1  s  surface  will  be  inversely  proportional 
to     the     square   of  its   distance  from   the   center   of  the  earth. 
Algebraically : 

W:w  ::d*  :  D2 

Suppose  we  wish  to  know  how  much  a  body  weighing  1,000 
Ib.  at  the  earth's  surface  will  weigh  1,000  miles  above  the 
surface.  Employing  the  above  proportion  we  have : 

1,000  :  w  ::  5,0002  :  4,0002,  from  which  w  =  640  Ib. 

A  body  below  the  earth's  surface  will  lose  in  weight  due  to 
the  fact  that  the  particles  above  the  body  exert  a  contrary 
attraction  to  that  of  the  particles  below.  The  resultant  force 
acting  on  the  body  will  be  the  difference  between  the  two 
attractions. 

As  the  polar  diameter  of  the  earth  is  less  than  the  equatorial 
diameter,  it  follows  that  gravity  is  greater  at  the  poles 
than  at  the  equator.  The  effect  of  gravity  is  also  lessened 
at  the  equator  due  to  the  centrifugal  force  set  up  by  the 
earth's  rotation.  If  the  earth  were  to  rotate  17  times  as 
fast  as  it  does,  the  force  of  gravity  at  the  equator  would  be 
neutralized. 

33.  The  Acceleration  of  Gravity. — Since  the  force  of  gravity 
varies   from   place   to   place,    the   acceleration   imparted   to 
falling  bodies  will  vary  slightly.     A  body  at  the  equator  will 
fall  with  less  velocity  than  a  body  at  the  poles.     The  accelera- 
tion of  gravity  is  the  velocity  a  freely  falling  body  will  gain 
during  each  second  of  fall  and  will  be  written  hereafter  as  "g." 
At  the  latitude  of  New  York  City  (40.73°N.),  the  acceleration 
of  gravity  is  32.16  ft.  per  second  per  second  or  980  cm.  per 
second  per  second. 

34.  Center  of  Gravity. — Since  all  bodies  are  composed  of 
small  particles,  it  naturally  follows  that  each  particle  is  acted 


26 


MECHANICS 


upon  by  gravity  individually.  Thus  we  may  consider  a 
body  as  acted  upon  by  parallel  forces,  the  sum  of  which  con- 
stitutes the  weight  of  the  body.  For  convenience,  we  may 
assume  that  all  of  the  separate  forces  may  be  replaced  by  a 
single  force  having  the  same  effect  as  the  joint  action  of  the 
separate  forces.  The  point  at  which  the  resultant  force 
would  be  applied  is  called  the  center  of  gravity  or  center  of 
mass  of  the  body.  The  center  of  gravity  is  the  paint  at  which 
the  weight  of  a  body  may  be  considered  concentrated.  If  a  body 
is  suspended  at  its  center  of  gravity,  there  will  be  no  tendency 
toward  rotation.  The  weight  of  a  body  is  always  taken  as  a 
separate  force  acting  downward  at  the  center  of  gravity. 


36.  Determination  of  the  Center  of 
Gravity. — In  regular,  homogeneous 
bodies,  the  center  of  gravity  may  be 
found  geometrically.  The  center  of 


C.6. 


FIG.  20. — Center  of  gravity  determined 
geometrically. 


(b) 

FIG.  21. — Center  of 
gravity  determined  ex- 
perimentally. 


gravity  of  a  uniform  lever  is  at  its  middle  point;  of  a  rectangle, 
the  intersection  of  the  diagonals;  of  a  circle,  the  center;  of 
a  triangle,  the  intersection  of  the  medians,  etc.  (See  Fig.  20.) 
If  the  body  is  irregular  in  shape,  the  center  of  gravity  may  be 
determined  experimentally.  For  example,  suppose  we  have  a 
piece  of  sheet  zinc  as  shown  in  Fig.  21  (a).  The  zinc  is  freely 
suspended  from  hole  "a"  and  a  plumb  line  dropped  from  the 
point  of  suspension.  The  center  of  gravity  will  be  along 
this  line  which  is  marked.  The  zinc  is  then  suspended 
from  hole  "6"  and  the  previous  operation  is  repeated.  The 


GRAVITATION  AND  GRAVITY  27 

center  of  gravity  will  be  at  the  intersection  of  the  two  posi- 
tions assumed  by  the  plumb  line.  If  a  hole  is  drilled  at  the 
point  of  intersection,  the  sheet  will  remain  fixed  in  any  posi- 
tion. The  center  of  gravity  of  levers,  rods,  etc.,  may  be  found 
by  balancing  over  a  knife  edge  (Fig.  21  (6)  ).  More  compli- 
cated determinations  of  the  center  of  gravity  are  omitted 
as  being  beyond  the  scope  of  this  book. 

36.  Equilibrium. — A  body  is  said  to  be  in  equilibrium  with 
respect  to  gravity  when  a  vertical  line  from  its  center  of 
gravity  passes  through  the  supporting  base.  If  the  vertical 
line  passes  outside  of  the  supporting  base,  a  rotation  results 
and  the  body  is  overturned.  There  are  three  kinds  of  equi- 
librium; (1)  stable;  (2)  unstable;  and  (3)  neutral. 


FIG.  22. — Bodies  A,  B  and  C  are  respectively  in  stable,  unstable  and  neutral 

equilibrium. 

The  three  kinds  of  equilibrium  are  illustrated  in  Fig.  22. 
A,  B  and  C  are  metal  disks  with  their  centers  of  gravity  as 
shown.  If  A  is  freely  suspended  at  point  PI,  it  is  said  to  be  in 
stable  equilibrium.  If  displaced,  the  body  tends  to  resume 
its  former  position,  due  to  the  fact  that  any  rotation  necessi- 
tates a  raising  of  its  center  of  gravity.  Since  B  is  suspended 
at  P2,  any  displacement  will  lower  the  center  of  gravity  and 
the  body  will  not  return  to  its  first  position.  B  is  therefore 
in  unstable  equilibrium.  C  is  suspended  at  PS.  Since  this 
point  coincides  with  the  center  of  gravity,  it  is  evident  that  a 
displacing  force  will  neither  raise  nor  lower  the  center  of 
gravity  and  that  the  body  will  stay  wherever  it  is  put.  C  is 
therefore  in  neutral  equilibrium. 


28  MECHANICS 

37.  Stability. — A  body  is  said  to  possess  stability  if  it  is 
difficult  to  overturn.  Stability  depends 
upon  the  size  of  the  supporting  base 
and  the  location  of  the  center  of  gravity. 
A  large  supporting  base  and  a  low  center 


10* 


of  gravity  give  a  body  great  stability.  A 
racing  yacht  must  be  well  ballasted  to 
keep  it  from  overturning.  Cargo  ships 
are  always  ballasted  when  crossing  the 
ocean  with  no  cargo  aboard.  The 
center  of  gravity  is  thus  kept  low. 
u~  "~"L  Tall  chimneys  are  not  highly  stable, 

F1G'  ^3;r~i?od?  A^~C~D  as  the  supporting   base   is    relatively 

has  5  ft.-lb.  of  stability.  J 

small  and  the  center  of  gravity  rela- 
tively high.  The  stability  of  a  body  is  measured  by  the 
amount  of  work  necessary  to  overturn  the  body.  It  is  found  by 
multiplying  the  weight  of  the  body  (concentrated  at  the 
center  of  gravity)  by  the  vertical  distance  through  which  the 
center  of  gravity  is  raised.  Thus  in  Fig.  23,  the  body  ABCD 
has  a  stability  of  5  foot-pounds. 

Questions  and  Problems 

1.  What  is  meant  by  universal  gravitation  ? 

2.  State  and  illustrate  the  law  of  universal  gravitation. 

3.  Distinguish  between  gravity  and  weight. 

4.  For  what  is  a  plumb  line  used?    Does  it  always  point  to  the  center 
of  mass  of  the  earth?     Explain. 

6.  Why  does  a  body  weigh  most  at  the  earth's  surface? 

6.  State  the  formula  for  finding  the  weight  of  bodies  above  the 
earth's  surface. 

7.  State  two  reasons  why  a  body  weighs  less  at  the  equator  than  at 
the  poles. 

8.  What  is  meant  by  the  acceleration  of  gravity?     What  factors 
affect  it?    What  is  the  acceleration  at  New  York  City? 

9.  Explain  fully  what  is  meant  by  the  center  of  gravity. 

10.  State  various  methods  of  determining  the  center  of  gravity. 

11.  Define  equilibrium.     Describe  the  three  kinds  of  equilibrium. 

12.  What  is  stability?    How  is  it  measured?     Give  examples  of  high 
and  low  stability. 


GRAVITATION  AND  GRAVITY  29 

13.  A  body  weighs  50  Ib.  at  the  earth's  surface.     How  much  will  it 
weigh  5,000  miles  above  the  surface? 

14.  A  body  weighs  50  Ib.  10,000  ft.  above  the  earth's  surface.     How 
much  will  it  weigh  at  the  surface? 

15.  A  body  weighs  100  Ib.  at  the  earth's  surface.     How  high  above  the 
surface  would  it  have  to  be  in  order  to  weigh  40  Ib.  ? 

16.  A  body  4  X  3  X  2  ft.  weighs  200  Ib.     Compute   its  stability  as 
it  rests  on  each  face. 


CHAPTER  V 
FORCES 

38.  Force. — Before  attempting  a  definition  of  force,  let  us 
examine  the  various  effects  which  a  force  may  produce.     It  is 
evident,  from  common  experience,  that  a  force  may: 

(a)  Cause  a  body  at  Test  to  move, 
(6)  Bring  a  moving  body  to  rest, 

(c)  Cause  a  body  to  increase  or  decrease  its  speed, 

(d)  Alter  the  course  of  a  moving  body, 

(e)  Cause  a  change  in  the  shape  of  the  body  by  stretching,  compressing, 
bending,  twisting  it,  etc, 

Cf)  Cause  certain  combinations  of  the  above  effects. 

No  better  or  simpler  illustrations  of  the  effects  of  forces 
can  be  had  than  from  our  national  game, — baseball.  Let  us 
assume  that  the  teams  are  on  the  field  and  that  the  batter  is 
"up."  The  ball,  which  is  at  rest  (zero  motion),  will  have 
velocity  imparted  to  it  as  the  pitcher  makes  his  delivery. 
If  the  batter  swings  and  misses,  the  ball  wiU  settle  to  rest  in 
the  catcher's  mitt.  In  this  case,  the  ball  is  stopped  both  by 
the  muscular  effort  of  the  catcher  and  the  interia  of  the  mitt. 
On  the  second  delivers,  the  batter  may  drive  a  ball  served  to 
him  slowly  with  such  tremendous  velocity  that  it  will  clear 
the  stand  or,  if  the  delivery  is  fast,  bunt  it  slowly  into  the 
infield,  in  either  case  changing  both  the  amount  and  direction 
of  the  original  motion.  In  addition,  the  impact  of  the  ball 
and  bat  tends  to  change  the  shape  of  the  ball. 

From  the  above  discussion,  it  is  evident  that  a  force  is 
anything  tending  to  cause  a  change  in  the  amount  or  direction  of 
the  motion  possessed  by  a  body  or  a  change  in  the  shape  of  the 
body  itself. 

39.  Action  and  Reaction. — Whenever  one  body  acts  upon 
another  body,   the  body  acted  upon  exerts  an  equal  and 

30 


FORCES  31 

opposite  effect  commonly  called  a  reaction.  The  following 
illustrations  will  make  clear  the  difference,  between  action  and 
reaction. 

1.   Suppose  we  have  a  book  weighing  2  Ib.  resting  upon  a  table  as  in  Fig. 
24.     It  will  be  seen  that  gravity  pulls 
the  book  against  the  table  with  a  force  -^ 

of  2  Ib.  It  will  be  seen,  also,  that  the 
table  pushes  up  against  the  book  with  a 
force  of  2  Ib.;  otherwise  the  book  would 
move  down  through  the  table.  Gravity  _ 


is    the    action,    and    the   upward   force 
exerted  by  the  table  is  the  reaction. 

2.  A  horse  is  tied  to  a  post  by  means  FIG.  24. — Action  and  reaction, 
of  a  halter.     If  the  horse  pulls  horizon- 
tally  with  a  force  of  200  Ib.,  the  post  must  exert  a  force  of  200  Ib. 
in  the  opposite  direction.     The  horse  causes  the  action  and  the  post  the 
reaction. 

3.  Two  boys  are  pulling  at  the  opposite  ends  of  a  rope.     If  one  boy 
exerts  a  force  of  40  Ib.,  the  other  boy  must  exert  a  force  of  40  Ib.  in  the 
opposite  direction.    In  this  case,  either  boy  may  be  considered  as  causing 
the  action. 

4.  An  automobile  is  being  towed  over  a  level  stretch  at  uniform  speed. 
The  force  exerted  by  the  tow  rope  is  the  action  and  friction  is  the  reaction. 

5.  If  the  automobile  in  (4)  is  being  towed  at  increasing  speed,  the  force 
exerted  by  the  tow  line  is  the  action,  and  friction  plus  inertia  is  the 
reaction. 

6.  If  the  automobile  in  (5)  is  being  towed  up  an  incline,  the  force 
exerted  by  the  tow  line  is  the  action;  and  friction  plus  inertia  plus  the 
the  effect  of  gravity  is  the  reaction. 

40.  Measurement  of  a  Force. — We  have  seen  that  a  force 
tends  to  change  the  amount  or  direction  of  the  motion  pos- 
sessed by  a  body  or  effect  a  change  in  the  shape  of  the  body. 
It  will  be  clear,  then,  that  the  amount  of  change  produced  will 
determine  the  magnitude  of  the  force. 

A  force  may  be  determined  in  magnitude: 

(a)  By  the  velocity  it  will  impart  to  a  given  body  in  a  given 
time. 

(b^  By  the  amount  of  change  of  shape  that  it  will  cause  a 
body  to  undergo. 

41.  Graphical    Representation    of    Forces. — In    order    to 
represent  a  force  graphically,  three  factors  must  be  considered: 


32  MECHANICS 

(1)  the  magnitude  of  the  force;  (2)  the  direction  in  which  the 
force  acts;  and  (3)  the  point  at  which  the  force  is  applied  to 
the  body.  This  is  readily  done,  in  a  quantitative  way,  by 
the  use  of  an  arrow  drawn  to  a  suitable  scale. 

For  example,  suppose  a  body  B  (Fig.  25)  to  be  acted  upon 
by  a  pulling  force  of  100  Ib.  (Fi)  acting  west  and  another 
pulling  force  of  200  Ib.  (F2)  acting  east.  Assuming  that  1  in. 
=  200  Ib.,  it  is  necessary  only  to  draw  a  line  K  in-  l°ng  due 
west  from  B,  and  another  line  1  in.  long  due  east  from  B. 

Scale.  !"=200#  Scale  •  I 


F,  - 


FIG.  25.  —  Forces  represented  graphi-          FIG.  26.  —  Forces  represented  graphi- 
cally. cally. 

The  arrows  show  the  direction  of  the  forces,  also  that  the  body 
B  is  being  subjected  to  tension,  tending  to  pull  apart  the 
particles  of  which  the  body  is  composed. 

In  case  FI  and  Fz  act  as  pushing  forces,  the  arrows  would  be 
placed  as  shown  in  Fig.  26.  The  body  B  is  now  under  com- 
pression, tending  to  crowd  the  particles  of  which  it  is  com- 
posed closer  together. 

42.  Moment  of  a  Force.  —  We  have  seen  that  forces  have  a 
tendency  to  produce  motion.  Under  certain  conditions,  the 

F_    #  motion  may  take  the  form  of 

„  *  A       rotation  about  a  fixed  point 

(fulcrum  or  axis  of  rotation)  . 

|  i  ~^~\5       For  example,  suppose  AB 

|<  .....  /2»*«--J  (Fig.  27)  to  be  a  rigid  bar 

without  weight,  arranged  to 
0  turn    about   a    pin    at   the 

fulcrum  (/)  and  acted  upon 

by  the  forces  FI  and  F2.  It  is  evident  that  FI  tends  to 
produce  a  clockwise  rotation  and  F2  a  counter-clockwise 
rotation.  Each  force,  then,  tends  to  set  up  an  opposite  rota- 
tion about/. 

The  measure  of  the  tendency  of  a  force  to  produce  rotation 
about  a  given  point  is  called  the  moment  of  that  force. 


FORCES  33 

Referring  again  to  Fig.  27: 

The  moment  of  FI  =  20  X  12  =  240  pound-inches. 

The  moment  of  F2  =  10  X  24  =  240  pound-inches. 

The  rotative  effect  of  each  force  is  the  same  and  the  bar  A  B 
will  be  at  rest.  Thus  a  small 
force  with  a  long  moment 
arm  may  produce  the  same 
results  as  a  large  force  with 
a  small  moment  arm. 

The  moment  arm  of  a  force     .  _^ ..      . 

A\  ^~,  -         ^-  \° 

is  the  perpendicular  distance 

from  the  line  in  which  the  ^,    /' 

force  acts  to  the  fulcrum  or 

point   of  rotation.     The 

Moment1  of  a  Force  =  the  FIG.  28. 

Force  X  the  Moment  Arm. 

In  the  case  of  forces  acting  at  an  angle,  it  is  necessary  to 
find  the  moment  arm  as  shown  in  Fig.  28.  Thus  the  moment 
arms  of  FI  and  Fz  are  di  and  dz  respectively.  A  determination 
of  the  perpendicular  distances  often  necessitates  an  extension 
of  the  force  lines.  The  extensions  should  always  be  dotted 
lines. 

Questions  and  Problems 

1.  State  the  various  effects  which  a  force  may  produce. 

2.  From  your  own  observations,  give  practical  illustrations  of  the 
above. 

3.  Define  force.     How  may  a  force  be  represented  graphically? 

4.  Give  examples  of  tension  and  compression. 

6.  State  two  ways  by  which  a  force  may  be  measured. 

6.  Under  what  conditions  will  a  force  produce  rotation? 

7.  What  is  meant  by  the  moment  of  a  force? 

8.  Define  moment  arm;  fulcrum;  torque. 

9.  State  clearly  how  you  would  find  the  moment  of  a  certain  force. 

10.  Explain  why  forces  never  exist  singly. 

11.  Distinguish  carefully  between  action  and  reaction,  giving  simple 
illustrations. 

12.  Indicate  the  actions  and  reactions  in  the  following  cases: 

1  For  pulleys,  etc.,  the  moment  is  often  called  torque. 

3 


34  MECHANICS 

(a)  A  team  of  horses  towing  an  automobile  on  the  level  at  constant 


(6)  A  team  of  horses  towing  an  automobile  up  a  hill  at  constant  speed. 

(c)  A  team  of  horses  towing  an  automobile  up  a  hill  at   increasing 
speed. 

(d)  Firing  a  gun. 

(e)  A  train  coasting  into  a  station. 

(/)  A  Hudson  River  steamboat  proceeding  north  at  increasing  speed; 
the  same  boat  going  down  stream  at  increasing  speed, 
(gr)  The  roof  truss  (Fig.  63).     Use  selected  points. 


CHAPTER  VI 
MOTION 

43.  Motion. — Motion  is  the  change  of  position  which  a  body 
undergoes  with  reference  to  some  given  point  or  object  and  is 
always   brought   about   by   the   application   of  force.     Rest 
means  a  permanence  of  position  with  reference  to  some  given 
point  or  object.     Rest  is  simply  a  statement  of  zero  velocity, 
indicating  lack  of  change  of  position  with  respect  to  some  point 
or  object. 

44.  Absolute  and  Relative  Motion. — If  we  wish  a  highly 
scientific  statement  of  motion,  we  must  record  the  successive 
changes  with  respect  to  some  ideal  point  fixed  in  space.     The 
motion  is  then  absolute. 

Ordinarily  we  think  of  motion  with  respect  to  some  nearby 
object  (or  objects),  in  which  case  the  motion  is  relative.  In 
defining  the  speed  of  an  automobile  as  30  m.p.h.,  the  ground 
is  taken  as  a  basis  of  comparison.  Actually  both  the  ground 
and  automobile  are  in  motion  on  account  of  the  earth's  rota- 
tion and  revolution.  Thirty  m.p.h.  simply  means  that  the 
speed  of  the  automobile  in  a  given  direction  exceeds  that  of  the 
earth  in  the  same  direction  by  30  miles  per  hour. 

Two  express  trains,  travelling  at  the  rate  of  60  m.p.h.  in 
the  same  direction,  have  a  velocity  of  zero  with  respect  to 
each  other,  and  a  velocity  of  60  m.p.h.  each  with  respect  to  the 
earth.  The  same  trains,  travelling  in  opposite  directions, 
have  a  velocity  of  60  m.p.h.  with  respect  to  the  earth  and  a 
velocity  of  120  m.p.h.  with  respect  to  each  other. 

A  man  walking  forward  in  a  submerging  submarine  will 
have  various  velocities.  He  will  have  a  velocity  with  respect 
to  the  land,  the  water,  members  of  the  crew,  a  torpedo  just 
discharged,  etc.  It  is  evident,  from  the  foregoing  illustra- 

35 


36  MECHANICS 

tions,  that  for  practical  purposes  all  motion  is  relative;  that  is, 
to  define  the  motion  of  any  body,  we  must  refer  it  to  some 
nearby  body  as  a  basis  of  comparison. 

45.  Translatory  and   Rotary  Motion. — In   order  to   have 
translatory  motion  (motion  of  translation),  every  particle  of 
which  a  rigid  body  is  composed  must  have  the  same  linear 
velocity  at  each  instant  as  every  other  particle.     It  is  evident 
that  each  particle  will  travel  in  a  path  either  parallel  to,  or 
coincident  with,  that  of  its  neighbor.     Furthermore,  a  straight 
line  connecting  any  two  particles  will  always  keep  the  same 
direction.     Translatory    motion    may    be    curvilinear    (in    a 
curved  line)  or  rectilinear  (in  a  straight  line).     It  may  also  be 
uniform  or  variable.     The  motion  of  an  engine  piston  and  a 
pile  driver  are  good  examples  of  rectilinear-translatory  motion. 
A  body  projected  horizontally  from  a  tower  without  rotation 
is  a  good  example  of  curvilinear-translatory  motion. 

In  rotary  motion  (motion  of  rotation),  each  particle  travels 
at  the  same  angular  velocity  as  the  other,  but  particles 
farthest  from  the  axis  of  rotation  will  possess  the  greatest 
linear  velocity.  It  will  be  seen  that  each  particle  will  describe 
a  course  either  parallel  to,  or  coincident  with,  that  of  its 
neighbor;  and  that  only  particles  equally  distant  from  the 
axis  will  travel  with  the  same  linear  velocity.  The  motion  of 
a  flywheel  and  an  airplane  propeller  are  common  examples. 

46.  Accelerated  Motion. — Acceleration  is  the  rate  at  which 
the  velocity  of  a  body  is  increased  or  decreased.     For  example,  if 
an  automobile,  starting  from  rest,  increases  its  velocity  2  ft. 
every  second  for  10  consecutive  seconds,  the  acceleration  is 
then  2  ft.  per  second  per  second  (written  2  ft./sec.2).     This 
simply  means  that  the  velocity  increases  at  the  rate  of  2  ft./ 
sec.   during  every  second.     The  same  acceleration  may  be 
expressed  as  120  ft./min./sec.     In  the  above  illustration  the 
acceleration  is  uniform  and  positive,  that  is  to  say,  the  change 
of  velocity  is  constant  and  causes  an  increase  in  the  speed  of 
the  automobile.     When  a  body  is  so  acted  upon  by  an  accelera- 
tion that  a  decrease  in  velocity  results,  the  acceleration  is 
negative.     For  example,  if  a  projectile  is  shot  vertically  upward 


MOTION  37 

with  a  velocity  say  of  1,000  ft.  per  sec.,  the  acceleration  of 
gravity  causes  the  body  to  lose  32.16  ft.  per  second  during  each 
second  and  will  eventually  stop  its  flight.  The  acceleration 
of  gravity  is  negative  for  rising  bodies  and  positive  for  falling 
bodies. 

47.  Speed  and  Velocity. — Speed  is  the  rate  at  which  a  body 
moves.  It  concerns  itself  with  displacement  and  time  only,  the 
direction  not  being  taken  into  consideration.  An  express 
train  travelling  at  the  rate  of  60  m.p.h.  is  said  to  have  a  speed 
of  60  m.p.h.  In  discussing  the  velocity  of  the  train,  we  must 
include  another  element, — direction.  If  the  train  in  proceed- 
ing due  north,  it  has  then  a  velocity  of  60  m.p.h.  due  north. 

In  case  the  velocity  of  a  body  is  variable,  it  is  expressed  as  if 
the  velocity  of  the  body  were  to  become  uniform  at  the 
particular  instant.  This  is  called  instantaneous  velocity.  A 
body  with  an  instantaneous  velocity  of  10  m.p.h.  will  not 
necessarily  cover  10  miles  in  an  hour.  The  distance  covered 
will  depend  on  the  average  velocity  for  the  given  time. 

The  velocity  (v)  at  the  end  of  any  second  for  a  uniformly 
accelerated  body  starting  from  rest  is  equal  to  the  accelera- 
tion (a)  multiplied  by  the  time  (t).  Thus  the  velocity  of  a 
freely  falling  body  at  the  end  of  the  first  5  seconds  will  be 
32.16  X  5  or  160.80  ft.  /sec. 

v  =  at 

If  a  body  is  uniformly  accelerated,  the  average  velocity 
(v  av.)  may  be  found  by  adding  the  initial  velocity  (v  in.) 
to  the  final  velocity  (v  fin.)  and  dividing  by  2.  To  illustrate, 
suppose  a  body  falls  from  a  state  of  rest.  Since  the  accelera- 
tion of  gravity  is  32.16  ft./sec.2,  the  velocity  at  the  end  of  3 
seconds  will  be  96.48  ft./sec.  The  average  velocity  for  the 
3  seconds  will  be  (0+96.48) /2  or  48.24  ft./sec. 

v  in.  -f  vfin. 
v  av.  -         -g- 

In  case  the  velocity  of  a  body  is  uniform,  the  total  distance 
(S)  covered  in  a  number  of  seconds  is  equal  to  the  uniform 
velocity  (v  un.)  multiplied  by  the  time  (t)  in  seconds.  For 


38  MECHANICS 

example,  if  a  body  has  a  uniform  velocity  of  1,000  ft. /sec.,  in 
10  seconds  it  will  travel,  1,000  X  10  or  10,000  ft. 

S  =  v  un.  X  t 

If  the  velocity  of  a  body  is  uniformly  accelerated,  the  total 
distance  (S)  covered  in  a  number  of  seconds  (t)  will  equal  the 
average  velocity  (v  av.)  multiplied  by  the  time.  Accordingly, 
a  freely  falling  body  will  cover  144.72  ft.  during  the  first 
3  seconds  of  fall.  The  average  velocity  for  the  3  seconds  will 
be  (0  +  96.48) /2  or  48.24  ft./sec.  48.24  X  3  =  144.72  ft. 
S  =  v  av.  X  t 

48.  Momentum. — The  quantity  of  motion  possessed  by  a 
moving  body  is  called  momentum.     Momentum  is  measured 
by  the  product  of  the  mass  and  velocity  of  the  body. 

Momentum  =  Mv 

There  is  no  recognized  unit  of  momentum  in  the  English 
system  of  measurement.  For  practical  purposes,  we  may 
assume  the  unit  to  be  a  mass  of  one  pound  moving  at  the  rate 
of  1  ft.  per  second.  For  example,  a  10,000  ton  ship  with  a 
speed  of  20  ft./sec.  is  said  to  have  400,000,000  f.p.s.  units  of 
momentum  (10,000  X  2,000  X  20). 

In  the  c.g.s.  system,  the  unit  of  momentum  is  the  bole. 
It  is  the  momentum  produced  when  a  mass  of  1  gram  is 
moving  at  the  rate  of  1  cm. /sec.  The  number  of  boles  is 
equal  to  the  product  of  the  mass  in  grams  and  the  velocity  in 
centimeters/sec,  (b  =  Mv).  For  example,  a  100  Kg.  pro- 
jectile with  a  velocity  of  300  m./sec.  will  have  a  momentum 
of  3,000,000,000  c.g.s.  units  or  3,000,000,000  boles  (100  X 
1,000  X  300  X  100). 

49.  Newton's  Laws  of  Motion. — The  following  laws  were 
formulated  by  Sir  Isaac  Newton.     A  thorough  understanding 
of  these  laws  is  very  important. 

I.  Every  body  continues  in  its  state  of  rest  or  of  uniform  motion 
in  a  straight  line,  unless  compelled  to  change  that  state  by  an 
impressed  force. 

II.  Change  of  momentum  is  proportional  to  the  force  acting 
and  takes  places  in  the  direction  in  which  the  force  acts. 


MOTION 


39 


///.  To  every  action  there  is  always  an  equal  and  opposite 
reaction. 

50.  Newton's  First  Law. — A  body  is  incapable  of  putting 
itself  in  motion.     A  book,  lying  on  a  table,  will  remain  in  the 
same  position  indefinitely,  unless  acted  upon  by  some  outside 
force.     Likewise,  a  body  in  motion  will  continue  to  move 
indefinitely,   unless  acted  upon  by  some  outside  force.     A 
thrown  baseball  tends  to  continue  on  forever  in  a  straight 
line.     Were  it  not  for  gravity  and  the  retarding  effect  of 'the 
atmosphere,  the  ball  would  never  come  to  rest.     It  is  clear 
from  this  discussion  that  Newton's  first  law  is  a  statement  of 
the  law  of  inertia. 

51.  Newton's  Second  Law. — It  is  obvious  from  Newton  s 
second  law  that  the  amount  of  change  of  momentum  undergone 
by  a  body  depends  on  the  magnitude  of  the  force  acting. 
Thus  we  are  able  to  measure  a  force  by  the  change  in  momen- 
tum it  will  produce  in  a  given  time.     The  force  varies  directly 
as  the   change  in  momentum  and  , __ 

u 

inversely  as  the  time  consumed.  ' a — r-. — ,— J 


F  a  Mv  /  t 

If  the  impressed  force  acts  counter 
to  the  motion  of  the  body,  the 
momentum  will  be  decreased;  if 
with  the  motion  of  the  body,  the 
momentum  will  be  increased. 

According  to  the  second  law,  a 
cannon  ball  dropped  from  a  height  FIG.  29. — A  and  B  will  reach 
of  200  ft.  will  strike  the  ground  at  the  f 
exactly  the  same  moment  as  another  ball  shot  horizontally 
from  the  same  height.  This  is  due  to  the  fact  that  each  ball 
is  given  the  same  downward  acceleration  by  gravity.  Further, 
the  balls  will  always  keep  the  same  relative  position  from  the 
ground  while  in  the  air.  We  see  also,  from  the  illustration, 
that  the  change  of  momentum  is  always  in  the  direction  in 
which  the  force  acts.  The  first  ball  is  acted  upon  by  gravity 
only  and  has  simply  a  downward  momentum,  The  second 


40  MECHANICS 

ball  is  acted  upon  by  the  force  of  the  explosion  as  well  as 
gravity  and  has  both  a  horizontal  and  downward  momentum, 
each  change  of  momentum  being  independent  of  the  other. 

A  simple  experiment  will  illustrate  the  above  law.  A  and 
B  (Fig.  29,  essential  details  shown  only)  are  two  similar  steel 
balls.  B  rests  on  the  support  F  and  A  is  held  in  place  by  the 
friction  between  C  and  G.  The  thumb  screw  D  enables 
proper  adjustment  of  the  spring  C  attached  at  H.  With  the 
balls  A  and  B  at  the  same  level,  the  hammer  E  is  allowed  to  fall 
against  C.  As  a  result  A  is  released  and  falls  to  the  floor, 
while  B  is  driven  out  horizontally  and  describes  a  parabolic 
path  to  the  floor.  Each  ball  reaches  the  floor  at  the  same 
instant. 

52.  Newton's  Third  Law. — Any  action  between  two  bodies 
is  mutual.  In  jumping  from  a  boat  to  the  shore,  the  boat 

tends  to  gain  an  equal  and 
opposite  momentum  to  that  of 
the  person  jumping.  A  per- 
fectly elastic  ball  thrown 


\ 


\ 


\ 


\  against  a  brick  wall  tends  to 

\— v,      bound    back    with    the    same 
f      \ 

(^       i     velocity  with  which  it  struck 
^gi       the  wall.     In  firing  big  guns  it 
^^  is  necessary  to  absorb  the  reac- 

FIG.  so.— Action  and  reaction.       tion  by  means  of  recoil  springs, 

otherwise  the  guns  might  break 

loose  from  their  foundations.     In  case  of  an  inelastic  body 
like  putty,  the  reaction  will  have  a  flattening  effect. 

In  Fig.  30,  A  and  B  are  two  highly  elastic  steel  balls.  If 
B  is  displaced  to  the  position  of  B'  and  allowed  to  fall  against 
A,  the  impact  will  cause  A  to  undergo  an  equal  displacement 
in  the  opposite  direction  as  shown  in  A'.  In  case  A  is  held 
fast,  B  will  bound  back  to  B' . 

53.  Graphical  Representation  of  Motion. — A  single  force, 
if  of  sufficient  magnitude,  will  cause  a  body  to  move  in  a 
straight  line.  The  motion  is  fully  defined  when  the  following 
are  given :  (a)  the  amount  of  the  motion;  (b)  the  direction  of  the 


MOTION 


41 


motion;  (c)  the  former  position  of  the  body.     To  represent  the 

motion  of  a  body  graphically,  it  is  only  necessary  to  draw  a 

straight  line  of  suitable  length,  with  an  arrow  head  pointing 

in  the  proper  direction.     Such  a  line  is  called  a  vector  or 

vector  line.     To  illustrate,  suppose  $ 

we  wish  to  represent  the  motion 

of  a  body  travelling  east  with  a 

velocity  of   10   ft./sec.     First   we 

adopt  some  convenient  scale,  such 

as    1    in.  =  2    ft./sec.     Then    we 

draw  a  line  5  in.   long,   using  an 

east  and  west  axis.     At  the  east 

extremity  we  draw  an  arrow  head 

pointing  due  east.     In  construct- 


SCALE: 


FIG.  31. — Velocities  represented 
graphically. 


ing  vector  lines,  the  scale  used  should  be  indicated  clearly. 

In  Fig.  31,  there  are  four  vector  lines  indicating  velocities 
from  a  common  point  in  four  different  directions.  Since  the 
scale  used  is  J  in.  =  10  ft./sec.,  it  is  evident,  by  a  measure- 
ment of  the  lines,  that  A  has  a  velocity  of  10  ft./sec.  due 
west;  B  a  velocity  of  20  ft./sec.  due  southeast;  C  a  velocity 
of  15  ft./sec.  due  east;  and  D  a  velocity  of  5  ft./sec.  due 
north. 

54.  Velocity  of  Rotation. — In  expressing  the  velocity  of 
a  rotating  body  two  methods  are  in  use :  (a)  the  length  of  arc 
described  by  a  selected  point  in  a  given  time ;  or  (6)  the  angular 
change  that  the  point  makes  from  the  axis 
of  rotation  in  a  given  time.  The  point 
selected  is  considered  to  be  on  the  rim  of 
the  rotating  body.  In  the  first  case,  it  is 
clear  that  the  arc  described  will  be  in  pro- 
portion to  the  distance  to  the  center  of 
rotation;  while,  in  the  second  case,  the 
angular  velocity  will  be  independent  of  the 
distance  from  the  center. 

Angular  velocity  may  be  expressed  as  revolutions  per  minute 
\p.m.)  or  degrees  per  minute  (d.p.m.).  In  the  latter  case  a 
lit  called  a  radian  is  used.  A  radian  is  an  angle  of  such 


FIG. 


32. — Angle 
a  radian. 


42  MECHANICS 

magnitude  that  the  arc  which  it  intercepts  on  the  circumference 
is  exactly  equal  to  the  radius  of  the  circle. 

In  Fig.  32,  angle  0  is  a  radian  since  the  arc  A B  is  just  equal 
in  length  to  the  radius  r.  As  the  circumference  of  a  circle 
equals  2irr,  it  will  be  seen  that  in  every  circumference  there 
are  2?r  radians  (Zirr/r).  Also,  since  there  are  360°  in  a  circle 
it  is  evident  that  a  radian  is  equal  to  57.3°  (360/27r).  Further 
the  linear  velocity  of  any  point  may  be  determined  by  multiplying 
the  velocity  in  radians  by  the  distance  from  that  point  to  the 
axis  of  rotation.  If  the  angular  velocity  in  Fig.  32  is  10 
radians/sec.,  the  linear  velocity  of  any  point  in  the  circum- 
ference will  be  30  ft./sec.  (10  X  3). 

Questions  and  Problems 

1.  Define    (a)   motion;    (6)   rest.     How  is  a  body  set  in   motion? 
Brought  to  rest? 

2.  What  is  (a)  absolute  motion?     (6)  relative  motion?     Discuss  fully. 

3.  State  the  difference  between  speed  and  velocity  and  give  an 
example  of  each. 

4.  What  is   variable  velocity?     Average   velocity?     Instantaneous 
velocity?     How  would  you  compute  the  actual  distance  passed  over  in  a 
given  time  by  a  body  whose  velocity  is  variable  ?     Uniformly  accelerated  ? 

5.  Define  and  give  an  example  of  (a)  translatory  motion;  (6)  rotary 
motion.     Give  an  instance  of  a  combination  of  the  two. 

6.  What  is  acceleration ?     Uniform  acceleration?     Variable  accelera- 
tion?    Positive    acceleration?      Negative    acceleration?     Illustrate    in 
each  case. 

7.  What  is  the  mathematical  relation  between  velocity,  uniform 
acceleration  and  the  time?     Between  total  distance  passed  over,  average 
velocity  and  the  time?     Between  total  distance  passed  over,  uniform 
velocity  and  the  time? 

8.  Define  momentum.     What  is  the  formula  for  momentum? 

9.  What  is  the  unit  of  momentum  in  the  f.p.s.  system?     The  c.g.s. 
system? 

10.  State  and  discuss  Newton's  first  law  of  motion. 

11.  State  and  discuss  Newton's  second  law  of  motion.     What  general 
relation  is  there  between  momentum,  the  force  producing  it  and  the  time? 
Describe  an  experiment  to  illustrate  the  second  law. 

12.  State  and  discuss  Newton's  third  law  of  motion.     Give  an  experi- 
ment to  show  that  action  and  reaction  are  equal  and  opposite  in  direction. 

13.  What  is  a  vector  or  vector  line?     How  may  the  velocity  of  a  body 
be  represented  graphically?     Give  an  example. 


MOTION  43 

14.  In  how  many  different  ways  may  the  velocity  of  a  rotating  body 
be  expressed?  What  is  a  radian?  How  many  radians  in  a  circum- 
ference? How  many  degrees  in  a  radian?  How  may  the  velocity  in 
radians  be  reduced  to  linear  velocity? 

16.  Represent  graphically  a  velocity  of  100  ft. /sec.  acting  north. 

16.  Represent  graphically  velocities  of  100  ft. /sec.,  80  ft. /sec.,  70  ft./ 
sec.  and  65  ft. /sec.  acting  north,  east,  south  and  southwest  respectively 
from  the  same  point. 

17.  An    automobile    flywheel    is  making  1,200  r.p.m.     Express  the 
velocity  in  degrees/min.;  in  radians/min.     If  the  wheel  is  1%   ft.  in 
diameter,  what  is  the  linear  velocity  of  a  point  on  the  circumference? 

18.  A  locomotive  is  turned  around  (front  to  back)  on  a  turntable  in 
4  minutes.     Express  the  angular  velocity  in  radians/sec.     In  degrees/ 
hour. 

19.  Two  gyroscopes  are  running  with  velocities  of  2,000  r.p.m.  and 
200  radians/sec,  respectively.     Which  has  the  greater  velocity?     By 
how  much?     Give  answer  in  radians. 

20.  A  circular  saw  is  running  at  the  rate  of  20,000  radians/min.     The 
distance  between  the  extremities  of  two  diametrically  opposed  teeth 
is  18  in.     How  far  does  the  point  of  a  tooth  travel  per  second?     How 
many  r.p.m.  does  the  saw  make? 

21.  Express  the  velocity  of  a  point  on  the  equator  hi  r.p.m.     In 
ft./sec.     In   radians/sec.     Assume  circumference  of  the  earth  to   be 
25,000  mi. 

22.  A  grinder  is  running  at  the  rate  of  3,500  r.p.m.     By  how  much  does 
this  vary  from  a  speed  of  200  radians/sec?     Give  answer  in  r.p.m. 


CHAPTER  VII 
COMPOSITION  OF  FORCES  AND  VELOCITIES 

55.  Resultant  of  Two  or  More  Forces. — Whenever  two  or 
more  forces  in  the  same  plane  are  applied  to  a  body  at  a 
fixed  point,  these  forces  may  be  replaced  by  a  single  force 
which  will  have  the  same  effect  on  the  body  as  the  joint  action 
of  the  original  forces.  This  single  force  is  known  as  the 
resultant  force.  The  process  of  determining  the  resultant  of 
two  or  more  concurrent  forces  is  known  as  the  composition 
of  forces.  The  separate  forces  are  called  components. 

For  the  sake  of  simplicity,  let  us  assume  that  we  are  dealing 
with  two  forces  only.  It  is  evident  that  the  two  forces  may 
act  (1)  in  the  same  straight  line  in  the  same  direction;  (2) 
in  the  same  straight  line  in  opposite  directions;  (3)  or  at  an 
angle  with  each  other. 

1.  If  two  forces  act  in  the  same  straight  line  and  in  the 
same  direction,  it  is  evident  that  the  resultant  is  equal  to 
their  sum.     For  example,  if  two  forces  of  10  and  20  Ib.  each 
are  applied  to  a  body  and  act  due  east,  they  may  be  replaced 
by  a  single  or  resultant  force  of  30  Ib.  applied  at  the  same  point 
and  acting  due  east. 

2.  If  two  forces  act  in  the  same  straight  line  and  in  opposite 
directions,  it  is  evident  that  the  resultant  is  equal  to  their 
difference.     For  example,  if  two  forces  of  10  and  20  Ib.  act 
east  and  west  respectively,  they  may  be  replaced  by  a  single 
or  resultant  force  of  10  Ib.  applied  at  the  same  point  and  acting 
due  west. 

3.  If  two  forces  act  upon  a  body  at  an  angle,  the  resultant 
is  equal  neither  to  their  sum  or  difference,  but  must  be  found 
by  means  of  the  parallelogram  law.     This  will  be  described 
in  the  following  paragraph. 

44 


COMPOSITION  OF  FORCES  AND  VELOCITIES 


45. 


56.  Parallelogram  of  Forces. — Suppose  it  is  desired  to  find 
the  resultant  of  two  forces  acting  at  an  angle  and  represented 
graphically  as  in  Fig.  33.  Using  the  force  lines  FI  and  F2 
as  adjacent  sides,  a  parallelogram  is  constructed  and  a  diagonal 

ft 


FIG.  33. — The  resultant  R  has  the  same  effect 

and  Ft. 


the  joint  action  of 


from  the  point  of  application  is  drawn.  The  diagonal  repre- 
sents the  resultant  both  in  amount  and  direction,  according 
to  the  scale  used.  Thus  the  single  force  R  will  have  the  same 
effect  as  the  joint  action  of  FI  and  Fz- 

67.  Experimental  Verification  of  the  Parallelogram  Law. — 
The  following  experiment  was  performed  by  the  author  as  a 


FIG.  34. — Apparatus  for  verifying  the  parallelogram  law. 

classroom  demonstration,  the  purpose  being  to  prove  that  the 
parallelogram  law  is  true.     A  weight  of  19.4  Ib.  was  suspended 


-46  MECHANICS 

before  the  blackboard  as  shown  in  Fig.  34,  care  being  taken 
that  there  was  no  friction  between  the  weight  and  the  board. 
The  balance  readings  recorded  were  10  and  15  Ib.  We  now 
have  three  concurrent  forces  in  equilibrium.  It  is  evident 
that  the  combined  effect  of  the  upward  forces  is  just  sufficient 
to  hold  up  the  suspended  weight.  In  other  words,  the 
resultant  of  the  two  upward  forces  must  be  equal  and  opposite 
in  direction  to  W.  If  this  is  confirmed  by  the  parallelogram 
law,  we  may  assume  the  law  to  be  correct. 

The  three  forces  were  located  on  the  blackboard  by  placing 
a  rectangular  block  along  the  cords  and  marking  with  a  sharp 
piece  of  chalk.  The  forces  (10  and  15  Ib.)  were  represented 
graphically,  the  scale  used  being  1  in.  =  1  Ib.  The  parallelo- 
gram was  completed  and  the  diagonal  drawn.  The  diagonal 
measured  19.4  in.  According  to  the  scale  assumed,  19.4  in. 
represents  19.4  Ib.  Since  the  resultant  is  found  to  be  equal 
and  opposite  in  direction  to  W,  we  may  be  certain  that  the 
parallelogram  law  is  correct.  W  is  here  known  as  the  equi- 
librant.  The  equilibrant  is  always  equal  in  magnitude  and 
opposite  in  direction  to  the  resultant. 

58.  Composition  of  Velocities. — Since  velocities  are  brought 
about  by  the  application  of  forces,  it  logically  follows  that  the 
resultant  of  two  velocities  may  be  determined  in  the  same  way 
as  the  resultant  of  two  forces.     It  is  thought  that  no  discussion 
will  be  necessary  to  enable  the  student  to  solve  such  simple 
problems  as  will  be  given  him. 

59.  Forces  and  Velocities  Exceeding  Two  in  Number. — In 
case  we  have  to  deal  with  more  than  two  forces  acting  at  an 
angle,  it  is  necessary  to  find  the  resultant  of  two  of  them; 
then  to  find  the  resultant  of  the  resultant  just  found  and  the 
next  force.     This  process  is  continued  until  each  force  has 
been  used.     The  same  method  is  followed  in  case  of  more  than 
two  velocities. 

60.  Directions    for    Graphical    Work. — Accurate    work    is 
essential  in  solving  problems  by  the  graphical  method.     Care- 
lessness is  inexcusable  and  results  in  a  waste  of  time.     The 
following  directions  and  suggestions  should  be  studied  carefully. 


COMPOSITION  OF  FORCES  AND  VELOCITIES  47 

1.  Graphical  diagrams  should  be  drawn  in  pencil.     A  fairly  hard  pencil 
with  a  sharp  point  gives  the  best  results. 

2.  If  the  paper  is  of  good  grade,  the  lines  may  be  inked  over  with 
india  ink.     Never  use  ordinary  ink. 

3.  Be  sure  that  the  ruler  used  has  a  straight  edge. 

4.  Use  a  compass  in  completing  the  parallelogram. 

5.  Be  sure  that  all  angles  are  correctly  represented. 

6.  Avoid  small  parallelograms.     It  is  suggested  that  one  half  of  an 
ordinary  sheet  of  paper  be  allowed.     This  will  tend  to  increase  accuracy. 

7.  Use  squared  paper  if  possible. 

8.  Label  each  diagram  fully,  being  sure  to  indicate  the  scale  used. 

Questions  and  Problems 

1.  What  is  meant  by  the  composition  of  forces? 

2.  Define  (a)  resultant;  (6)  component;  (c)  equilibrant. 

3.  How  would  you  find  the  resultant  of  several  forces  acting  in  different 
directions  and  attached  at  the  same  point? 

4.  Find  the  resultant  of  the  forces  given  below: 
(a)  10  Ib.  and-  30  lb.;  angle  between   =    25°. 
(6)   100  lb.  and  60  lb.;  angle  between  =    40°. 

(c)  24  Kg.  and  36  Kg.;  angle  between  =    90°. 

(d)  100  lb.  and  200  lb.;  angle  between  =  125°. 

(e)  200  lb.  and  112  lb.;  angle  between  =  140°. 
(/)  500  g.  and  590  g.;  angle  between  =  100°. 

6.  A  motor  boat  has  a  vekx  ity  of  10  m.p.b.  upstream  (north).  The 
wind  gives  the  boat  a  velocity  of  2  m.p.h.  to  the  east.  Find  the  direction 
in  which  the  boat  goes  and  the  actual  distance  covered  in  one  hour. 

6.  An  automobile  is  travelling  north  at  a  rate  of  40  m.p.h.     The  wind 
is  blowing  from  the  west  at  the  rate  of  15  m.p.h.     What  is  the  resultant 
velocity  of  the  wind?     From  what  direction  does  the  wind  seem  to 
come? 

7.  In  what  direction  would  a  weather  vane  point  in  problem  6,  if 
mounted  on  the  radiator  of  the  car? 

8.  A  balloon  rises  at  the  rate  of  25  ft.  per  second.     It  is  blown  east  at 
the  rate  of  5  ft.  per  second.     Find  the  rate  and  direction  of  motion. 


CHAPTER  VIII 


RESOLUTION  OF  FORCES  AND  VELOCITIES 

61.  Resolution  of  Forces  and  Velocities. — It  has  been 
shown,  in  the  previous  chapter,  that  we  may  replace  two  or 
more  forces  (or  velocities)  applied  at  the  same  point  and  in 
the  same  plane,  by  a  single  force  (or  velocity)  having  the 
same  effect.  We  shall  now  see  that  it  is  possible  to  replace 
a  single  force  (or  velocity)  with  two  or  more  components 
having  the  same  joint  effect  as  the 
original  force  (or  velocity) .  The  process 
of  replacing  a  single  force  (or  velocity) 
with  two  or  more  forces  (or  velocities) 
having  the  same  joint  effect  as  the  original 
force  (or  velocity),  is  called  the  resolution 
of  forces  (or  velocities) .  Resolution  is 
the  opposite  of  composition.  Problems 
dealing  with  resolution  may  be  solved 
by  means  of  the  parallelogram  law. 

Suppose  we  wish  to  resolve  a  force 
FIG.  35.— F  may  be  re-   (F)  into  two  components  (Ci  and  £2) 

placed  by  its    components    ftt  leg   of  3QO    and    4Qo    respectively 

Ci  and  C2. 

with  the  force.     First  F  is  represented 

graphically,  both  in  amount  and  direction,  as  shown  in  Fig. 
35.  Ci  and  C2  are  drawn  from  0  of  indefinite  length  and  at 
the  proper  angles.  The  components  are  now  represented  in 
direction,  but  not  in  magnitude.  A  parallelogram  is  next 
formed  with  F  as  the  diagonal.  Thus  the  components  (Ci  and 
€2)  are  defined  in  length.  According  to  the  scale  assumed 
for  F,  they  are  now  determined  in  amount.  C\  and  Cz,  acting 
jointly,  will  have  exactly  the  same  effect  at  0  as  F. 

In  case  the  components  are  known  in  magnitude,  but  not  in 

48 


RESOLUTION  OF  FORCES  AND  VELOCITIES 


49 


direction,  a  pair  of  compasses  will  be  necessary  to  complete  the 
geometrical  construction. 

If  the  angles  are  decreased,  the  components  will  decrease; 
if  the  angles  are  increased,  the  components  will  increase. 

62.  Vertical  and  Horizontal  Components. — Frequently  it  is 
necessary  to  resolve  a  given  force  or  velocity  into  rectangular 
components.  Figure  36  represents  a  force  (F)  resolved 
graphically  into  vertical  and  horizontal  components  (V  and 
H).  The  vertical  component  (V)  represents  the  effective 

F 


FIG.  36. — Rectangular  components. 

value  of  F  vertically  and  the  horizontal  component  (H)  repre- 
sents the  effective  value  of  F  horizontally.  Thus  we  are  able 
to  determine  both  the  lifting  effect  and  the  horizontal  pull 
produced  by  F. 

63.  Value   of   Squared   Paper. — Graphical   results   in  the 
resolution  of  forces  and  velocities  are  facilitated  by  the  use  of 


Scale 


2x 


120* 


32Spaces 
FIG.  37. — Resolution  by  means  of  squared  paper. 

squared  paper.     For  example,  suppose  (Fig.  37)  we  have  a 
force  of  120  Ib.  acting  <£  degrees  to  the  horizontal.     If  the 


50  MECHANICS 

paper  is  ruled  in  J^o  in-  squares,  then  J^o  in.  =  3  Ib.  is  a  good 
working  scale.  Thus  to  represent  120  Ib.,  we  draw  a  line  2  in. 
long.  The  rectangular  components  may  now  be  determined  in 
value  directly,  a  rule  being  unnecessary.  By  inspection  we 
see  that  V  is  24  spaces  long  and  =  24  X  3  =  72  Ib.  in  magni- 
tude. H  is  found  to  be  32  spaces  long  and  =  32  X  3  =  96  Ib. 
in  magnitude. 

64.  Resolution  by  the  Trigonometrical  Method.  —  A  force 
or  velocity  may  readily  be  resolved  into  vertical  and  horizontal 

p  components  by  the  use  of  simple 
trigonometric  functions.  Refer- 
ring to  Fig.  38,  it  is  evident  that 
cos  30°  =  H/F,  orH=F  cos 
30°,  H  =  60  X  .866  =  51.96  Ib. 
Also  cos  60°  =  V/F,  or  V  =  F 
cos  60°,  V  =  60  X  .500  =  30  Ib. 

FIG.  38.  —  Rectangular  components         65.    Composition  of  Forces  by 
determined  trigonometrically.         Resolution 


Horizontal  Components.  —  If  several  forces  act  in  the  same 
plane  and  are  applied  at  the  same  point,  their  resultant  may 
be  found  by  resolving  all  the  forces  not  vertical  or  horizontal 
into  their  vertical  and  horizontal  components.  Since  the 
forces  are  now  all  vertical  or  horizontal,  by  proper  additions 
and  subtractions  we  may  combine  them  into  one  vertical  and 
one  horizontal  force.  The  resultant  of  the  last-mentioned 
forces  will  given  the  resultant  of  the  original  forces  both  in 
magnitude  and  direction. 

Questions  and  Problems 

(Solve  problems  graphically  unless  otherwise  instructed) 
'1.  Explain  what  is  meant  by  resolution  of  forces. 

2.  What  are  rectangular  components? 

3.  How  would  you  find  the  resultant  of  several  forces  applied  at  the 
same  point  and  in  the  same  plane? 

4.  Resolve  a  force  of  20  Kg.  into  two  components  at  angles  of  15°  and 
25°  with  the  force, 

5.  Resolve  a  force  of  200  Ib.  into  two  components  such  that  one  com- 
ponent will  be  a  force  of  120  Ib.  acting  at  an  angle  of  30°  with  the  200 
Ib,  force,    Give  also  the  angle  at  which  the  second  component  acts% 


RESOLUTION  OF  FORCES  AND  VELOCITIES 


51 


6.  Resolve  a  force  of  40  Ib.  into  two  components  of  20  Ib.  and  30 
Ib.  each  and  find  the  angle  between  each  component  and  the  original  force. 

7.  Resolve  a  force  of  500  Ib.  acting  at  an  angle  of  35°  to  the  horizontal 
into  its  vertical  and  horizontal  components.     Solve  with  squared  paper. 

8.  Repeat  Problem  7,  using  a  trigonometrical  solution. 

9.  Repeat  as  in  Problems  7  and  8  for  a  force  of  1,000  Ib.  acting  at 
an  angle  of  60°  with  the  horizontal. 

10.  A  tug  boat  is  towing  a  scow  at  an  angle  of  5°  with  the  axis  of  the 
scow.     If  the  tension  in  the  tow  line  is  300  Ib.,  what  force  is  moving  the 
scow  a  head  and  what  force  is  tending  to  move  the  scow  to  the  side? 

11.  Determine  the  resultant  of  the  sets  of  forces  in  Fig.  39.     Use  the 
method  of  resolution  into  rectangular  components  in  (2)  and  (3). 


5* 


W<T  6^r 


(1) 


&50* 

V 


(2) 
FIG.  39. 


12.  Resolve  a  force  of  300  Kg.  into  two  components  acting  at  an  angle 
of  80°  with  each  other,  such  that  one  force  shall  be  three  times  the  other. 

13.  What  force  will  be  necessary  to  keep  a  100  Ib.  ball  on  an  inclined 
plane  10  ft.  long  and  3  ft.  high?     Assume  the  supporting  force  to  act 
parallel  to  the  plane  and  at  the  center  of  the  ball. 


CHAPTER  IX 
EQUILIBRIUM  OF  CONCURRENT  FORCES 

66.  Concurrent  Forces  in  Equilibrium. — Forces  applied  at 
the  same  point,  or  passing  through  the  same  point  if  extended, 

are  called  concurrent  forces. 
, '  Concurrent  forces  will  be  in 

equilibrium  (of  rest  or 
motion)  when  the  opposite 
vertical  forces  are  balanced 
""^     and  when  the  opposite  hori- 
zontal forces  are  balanced. 
Stated   differently,  a  body 
acted    upon    by    concurrent 
forces    will    be    in   equilib- 
FIG.  40.— Concurrent  forces  in  rium,  when  the  resultant  of 

equilibrium.  T7  ,7       /.  , . 

all  the  forces  acting  is  zero. 

For  example,  the  body  B  (Fig.  40)  is  at  rest.     It  is  evident, 
then,  that: 

7i  =  72  and  H1  =  #2 

A  body  acted  upon  by  a  single  force  or  by  unbalanced 
forces  can  never  be  in  equilibrium.  In  case  the  forces  are 
not  vertical  or  horizontal,  the  body  acted  upon  will  be  in 
equilibrium  if  the  resultant  of  the  vertical  components  is 
zero  and  the  resultant  of  the  horizontal  components  is  zero. 

Another  case  of  equilibriun  is  illustrated  in  Fig.  41.  A 
body  (B)  weighing  125  Ib.  is  being  moved  at  a  uniform  velocity 
along  the  surface  S-S1  by  a  force  of  50  Ib.  acting  at 
an  angle  of  25°  to  the  horizontal.  Suppose  we  wish  to  find 
the  force  of  friction  (Fr.)  and  the  perpendicular  pressure 
(P)  between  the  body  and  the  surface.  Referring  to  the 
"free  body"  diagram  (Fig.  42),  it  is  evident,  since  the 
body  is  in  equilibrium,  that: 

52 


EQUILIBRIUM  OF  CONCURRENT  FORCES 


53 


(1)  H  =  Fr.  and  (2)  V  +  P  =  G. 

Since  H  =  50  cos  25°,  and  V  =  50  cos  65°,  it  follows  that: 
(1)  50  cos  25°  =  Fr.,  and  Fr.  =  50  X  .906  =  45.3  lb., 

P 


125* 


FIG.  41.  FIG.  42. — Force  diagram  for  Fig.  41. 

(2)  50   cos   65°  +  P  =  125,    and   P  =  125  -  (50  X  .423)  = 
103.85  lb. 
67.  Triangle  of  Forces. — If  three  concurrent  forces  in  the 

D 


A; 


W 
FIG.  43. 


W 

FIG.  44. 


same  plane  are  in  equilibrium,  they  may  be  represented  in 
amount  and  direction  by  the  sides  of  a  triangle. 

Suppose  Fi,  F2  and  W  (Fig.  43)  to  be  in  equilibrium.     Then 


54 


MECHANICS 


as  shown  in  Fig.  44,  W  is  equal  to  R,  the  resultant  of  FI  and 
F2.  Still  referring  to  Fig.  44,  it  is  evident  that  FI,  F2  and 
W  are  proportional  respectively  to  the  sides  of  the  triangle 
ABD,  since:  (1)  FI  is  proportional  to  BA;  (2)  F2  is  propor- 
tional to  BC  and  therefore  to  AD;  (3)  W  =  R  and  is  therefore 
proportional  to  BD,  but  opposite  in  direction.  Hence  the 
original  forces,  FI,  Fz  and  W  are  represented  both  in  amount 
and  direction  by  the  sides  of  the  triangle  BA,  AD  and  DB,  as 
shown  in  Fig.  45. 

68.  Concurrent  Forces  Applied  at  Different  Points. — Con- 
current forces  may  be  applied  to  a  body  at  the  same  point 
or  they  may  be  applied  at  different  points.  In  the  latter 
case,  the  force  lines  will  meet  at  a  common  point  if  extended. 


FIG.  46.— Weight  of  bar  A-B  is  equal 
to  the  resultant  of  Fi  and  F2. 


FIG.  47. — Force  diagram 
for  Fig.  46. 


For  example,  AB  (Fig.  46)  is  a  rigid,  uniform  bar  weighing 
W  Ib.  and  supported  by  cords  and  balances  as  shown.  The 
weight  (W)  of  the  bar  acts  vertically  downward  at  the  center 
of  gravity  (C.G.).  W  is  balanced  by  the  tensions  (Fi  and  F2) 
of  the  supporting  cords.  It  is  clear,  since  the  bar  is  in  equilib- 
rium, that  W  is  numerically  the  same  as  the  resultant  of  FI 
and  F2.  The  plumb  line  (P),  suspended  at  C,  determines  the 
direction  of  W  and  will  always  pass  through  the  center  of 
gravity.  If  FI  and  Ft  are  known,  W  may  be  determined; 


EQUILIBRIUM  OF  CONCURRENT  FORCES 


55 


if  W  is  known,  FI  and  F2  may  be  determined.  The  force 
diagram  for  Fig.  46  is  shown  in  Fig.  47.  FI  and  F2  (balance 
readings)  are  drawn  to  scale  at  the  proper  angle  and  a  paral- 
lelogram formed.  The  resultant  (R)  is  found,  using  the  same 
scale,  and  will  check  with  the  weight  of  the  bar. 

In  case  the  supporting  cords  diverge,  as  in  Fig.  48,  the 
solution  is  similar  to  the  one  just  described.     W  is  numerically 

'    Q 


FIG.  48. — Weight  of  bar  A-B  is  equal  to  the  resultant  of  Fi  and  Ft. 

the  same  as  the  resultant  (R)  of  FI  and  F2.  The  dotted 
parallelogram  lines  will  always  intersect  along  the  plumb 
line  (P).  Figure  48  is  a  combined  picture  and  force  diagram. 
In  case  W  is  known  and  it  is  desired  to  find  FI  and  F2,  the 
force  diagram  will  be  the  same,  the  only  difference  being  that 
W  is  drawn  to  scale,  the  parallelogram  completed,  and  FI 
and  F2  computed.  The  computed  values  should  check  with 
the  balance  readings. 

Questions  and  Problems 

1.  What  are  concurrent  forces? 

2.  Under  what  conditions  will  concurrent  forces  produce  equilibrium? 


56  MECHANICS 

3.  What  is  meant  by  the  triangle  of  forces?     Explain  fully,  using 
diagram. 

4.  Describe  a  typical  experiment  involving  concurrent  forces  applied 
at  separate  points. 

6.  Find  the  amount  and  direction  of  a  force  that  will  produce  equilib- 
rium with  two  forces  of  30  and  50  g.  acting  at  an  angle  of  40°  with  each 
other. 

6.  A  block  weighing  5  Ib.  is  being  pulled  uniformly  along  a  surface 
by  a  horizontal  force  of  1  Ib.     Show  by  diagram  the  actions  and  reactions. 

7.  A  body  weighing  50  Ib.  is  being  pulled  uniformly  by  a  force  of 
12  Ib.  acting  at  an  angle  of  15°  to  the  horizontal  (See  Fig.  41).     Find 
the  perpendicular  reaction  of  the  surface  and  the  horizontal  pull. 

8.  A   picture  weighing  10  Kg.  is  suspended  by  two  cords  of  equal 
length  acting  at  an  angle  of  50°  with  each  other.     Find  tension  in  the 
cords. 

9.  A  wire  AB  30  ft.  long  is  fastened  at  either  end.     If  a  man  weighing 
160  Ib.  stands  in  the  middle  and  the  wire  is  depressed  2  ft.,  what  is 
the  tension  at  each  end? 

10.  A  bar  weighing  10  Ib.  is  suspended  as  in  Fig.  46.     Find  FI  and  F%. 


CHAPTER  X 
EQUILIBRIUM  OF  PARALLEL  FORCES 

69.  Tendencies  of  Parallel  Forces. — Figure  49  represents 
a  rigid,  weightless  body  acted  upon  by  parallel  forces.  Various 
effects  may  be  produced  by  the  forces,  according  to  their 
magnitudes  and  points  of  application.  They  may:  (a)  cause 
an  upward  displacement  of  the  body  AB'-(b)-  cause  a  down- 
ward displacement  of  AB;  (c)  cause  AB  to  rotate;  (d)  cause  a 

f. 


FIG.  49. — Body  acted  upon  by  parallel  forces. 

simultaneous  displacement  and  rotation. 

70.  Equilibrium  of  Parallel  Forces. — In  order  to  avoid  the 
effects  as  stated  in  the  previous  paragraph,  that  is,  in  order 
that  the  body  A B  shall  be  in  equilibrium  (at  rest),  the  follow- 
ing conditions  must  prevail : 

1.  The  sum  of  the  "up"  forces  must  equal  the  sum  of  the 
"down"    forces, 

FI  H-  F4  must  equal  F2  +  t\ 

2.  The  sum  of  the  moments  tending  to  produce  rotation  in 
one  direction  must  equal  the  sum  of  the  moments  tending  to 
produce  rotation  in  the  opposite  direction.     Assuming  any 
point  as  the  axis  of  rotation  (fulcrum),  e.g.,  point  A,  then, 

F2  X  (di  +  dj)  +  ^3  X  (di  +  d2  +  da)  must  equal 

Fi  X  d,  +  F4  X  (di  +  dz  +  d,  +  d,) 
57 


58 


MECHANICS 


Summarizing,  we  may  say  that  a  body  acted  upon  by 
parallel  forces  will  be  in  equilibrium  when: 

1.  The  sum  of  the  forces  acting  in  one  direction  equals  the 
sum  of  the  forces  acting  in  the  opposite  direction; 

2.  The  sum  of  the  clockwise  moments  equals  the  sum  of  the 
counter-clockwise  moments. 

71.  The  Laws  of  Parallel  Forces  Proved  by  Experiment. — 
The  conditions  necessary  for  equilibrium  as  stated  in  the 
previous  paragraph  may  be  verified  in  the  laboratory.  The 
following  experiment  was  performed  by  two  students. 


Ac 


\B C 


n of Bar 


*, 
FIG.  50. — Apparatus  for  studying  the  laws  of  parallel  forces. 

15*  f3.PS* 


B 


J<. 


<- 17' * 

11.25*  6.25*  10.75* 

FIG.  51. — Force  diagram  for  Fig.  50. 


AG  (Fig.  50)  is  a  rigid  steel  bar  weighing  6.25  Ib.  and  sup- 
ported by  two  spring  balances.  Weights  of  11.25  Ib.  and  10.75 
Ib.  were  suspended  from  C  and  E.  The  weight  of  the  bar  con- 


EQUILIBRIUM  OF  PARALLEL  FORCES 


59 


centrated  at  its  center  of  gravity  is  represented  by  a  force  of 
6.25  Ib.  acting  downward  at  D.  The  apparatus  was  carefully 
adjusted  until  AG  was  horizontal  and  the  balances  vertical. 
The  balance  readings  were  observed  and  the  distances  between 
the  forces  measured.  A  force  diagram  was  then  constructed, 
with  the  forces  and  distances  clearly  labelled  (Fig.  51). 

The  moments  of  the  various  forces  were  determined  first 
about  A  and  then  about  D.  The  results  were  finally  tabulated 
as  shown  below. 


Forces  in  Ib. 

Moments  around  A 
in  Ib.-in. 

Moments  around  D 
in  Ib.-in. 

At 

Up 

Down 

Clockwise 

Counter- 
clockwise 

Clockwise 

Counter- 
clockwise 

B 
C 
D 
E 
F 

15.00 

30.00 

330.00 

11.25 
6.25 
10.75 

78.75 
150.00 
397.75 

191.25 

13.25 

596.25 

139.75 

278.25 

Totals  .  .  . 

28.25 

28.25 

626.50 

626.25 

469.75 

469.50 

An  examination  of  the  above  tabulation  shows  that  the 
sum  of  the  "up"  forces  exactly  equals  the  sum  of  the  "down" 
forces;  also  that  the  sum  of  the  clockwise  moments  is  practic- 
ally the  same  as  the  sum  of  the  counter-clockwise  moments. 
The  small  difference  in  the  moments  is  due  to  the  fact  that  it 
is  impossible  to  measure  distances  with  100  per  cent,  accuracy. 
This  experiment  verifies  the  laws  of  parallel  forces  as  stated 
in  Par.  70. 

72.  The  Resultant  and  Its  Point  of  Application  for  a  System  of 
Parallel  Forces. — Suppose  it  is  desired  to  replace  the  system 
of  forces  in  Fig.  52  by  a  single  force  (resultant)  having  the 
same  effect  as  the  joint  action  of  the  separate  forces.  We  must 
find  the  magnitude  of  the  force  and  its  point  of  application. 
It  is  evident  that  the  resultant  will  be  equal  and  opposite 


60 


MECHANICS 


ft 

'*            E          10* 

i 

(3')           (5')    \     (7 

II 
1 
V     ' 
?#               20*     R    2. 
FIG.  52. 

P        4* 

in  direction  to  the  force  needed  to  cause  equilibrium.     This 
force  is  found  as  follows: 

15  +  E  +  10  =  10  +  20  +  25  +  45, 
E  =  75  Ib.  resultant  force. 

To  find  the  point  of  application,  assume  A  as  the  point  of 

rotation   and  take   distances 
from  A  as  in  the  parentheses. 
From  the  law  of  equilibrium 
of  parallel  forces  we  have : 

15  X  2  +  75  X  d  +  10  X  6 
=  20  X  3  +  25  X  5  +  45^ 
X  7,  d  =  5.46ft.  from  A. 

Hence  we  find  that  the  re- 
sultant of  the  above  forces  is 
a  down  force  of  75  Ib.  and  that  the  point  of  application  is 
5.46ft.  from  A. 

73.  Couple    Defined. — Two    equal    and    opposite    parallel 
forces  acting  at  separate  points  constitute  a  couple.     Refer- 
ring to  Fig.  53,  it  will  be  seen  that 
a    couple   causes   rotation   only   and 
that  there  can  be  no  resultant.     In 
order  to  produce  equilibrium,  a  second 
couple  having  an  equal  and  opposite  I 
effect  is  necessary.     The  distance  (d) 
between  the  forces  is  called  the  couple  J 
arm.     The  actual  couple  is  F\  X  d  or 

P     y  j  FIG.  53. — Couple. 

L    2     ^\    W'* 

Questions  and  Problems 

1.  State  the  conditions  necessary  for  equilibrium  of  parallel  forces. 

2.  Describe  a  laboratory  experiment  to  verify  the  above  conditions. 

3.  What  is  meant  by  the  resultant  df  a  system  of  parallel  forces? 

4.  What  is  a  couple?     Couple  arm?     What  kind  of  motion  does  a 
couple  produce? 

5.  A  painter's  scaffold  AB  is  20  ft.  long  and  weighs  100  Ib.     The 
painter,  weighing  150  Ib.,  is  standing  in  the  middle  of  the  scaffold.     Find 
the  tension  in  each  supporting  rope. 


k- 


EQUILIBRIUM  OF  PARALLEL  FORCES  61 

6.  Repeat  problem  5,  if  the  painter  is  standing  5  ft.  from  A. 

7.  Two  men  are  carrying  a  load  of  125  Ib.  on  a  uniform  pole  20  ft. 
long.     If  the  pole  weighs  25  Ib.,  where  will  the  load  have  to  be  placed 
in  order  that  one  man  shall  carry  twice  as  much  as  the  other?     Assume 
that  the  force  exerted  by  each  man  acts  1  ft.  from  the  end  of  the  pole. 

8.  A  light  and  a  heavy  horse  are  harnessed  together.     Where  must  the 
king  pin  in  the  evener  be  placed  in  order  that  the  light  horse  shall  pull 
only  %  as  much  as  the  heavy  horse? 

9.  It  is  customary  to  use  at  least  three  horses  on  a  reaper.     Design 
a  whiffletree  for  three  horses,  so  that  each  horse  shall  exert  the  same 
pull. 

10.  A  uniform  bridge  100  ft.  long  is  supported  by  abutments  A  and  B. 
The  bridge  weighs  175  tons  and  is  carrying  a  locomotive  weighing  60 
tons.     If  the  center  of  gravity  of  the  locomotive  is  30  ft.  from  A,  find 
the  vertical  reaction  of  each  abutment. 

11.  Repeat  Problem  10,  if  the  center  of  gravity  of  the  locomotive  is 
40  ft.  from  B. 

12.  A  uniform  bar  AB  is  10  ft.  long  and  weighs  20  Ib.     Three  forces  of 
8  Ib.  each  act  directly  down  on  the  bar  which  is  horizontal.     Two  of 
the  forces  act  at  the  extremity  of  the  bar  and  the  other  4  ft.  from  A. 
Find  the  resultant  and  point  of  application  of  the  forces. 

13.  Find  the  resultant  and  point  of  application,  if  the  forces  in  Fig. 
52  are  doubled  and  the  distances  remain  the  same. 

14.  A  door  7  ft.  high  and  3.5  ft.  wide  is  supported  by  hinges  1  ft. 
from  the  upper  and  lower  extremities.     The  door  weighs  50  Ib.     Assum- 
ing each  hinge  to  bear  an  equal  load,  what  is  the  vertical  and  horizontal 
reaction  at  each  hinge? 

15.  The  lever  of  the  safety  valve  shown  in  Fig.  54  is  29  in.  long.     The 
lever  weighs  3  Ib.  and  its  center  of  gravity  is  14  in.  from  the  fulcrum 


t 


FIG.  54. 


A.  The  valve  is  2  in.  in  diameter.  The  valve  stem  is  attached  to  the 
lever  3.5  in.  from  A.  Weight  of  the  valve  and  stem  is  2  Ib.  A  ball 
weighing  25  Ib.  is  suspended  from  B.  Make  a  free  body  diagram  from  the 
above  figures  and  compute  the  force  in  Ib.  per  sq.  in.  necessary  to  "blow" 
the  valve. 


CHAPTER  XI 
EQUILIBRIUM  OF  NON-CONCURRENT  FORCES 

74.  Non-concurrent  Forces  in  Equilibrium. — Forces  which 
do  not  pass  through  a  common  point  and  which  are  not  all 
parallel,    are    called    non-concurrent  forces.    Since    non-con- 
current forces  may  produce  displacement  as  well  as  rotation, 
it  is  evident  that  such  a  system  may  not  always  be  replaced 
by  a  single  force  or  resultant. 

Non-concurrent  forces  will  be  in  equilibrium  when : 

1.  The  resultant  of  the  vertical  forces  is  zero, 

2.  The  resultant  of  the  horizontal  forces  is  zero, 

3.  The  sum  of  the  clockwise  moments  is  equal  to  the  sum  of  the 
counter-clockwise  moments. 

If  any  of  the  three  conditions  stated  above  is  not  satisfied, 
there  can  be  no  equilibrium.  The  student  should  note  that 
if  the  original  forces  are  neither  vertical  nor  horizontal,  the 
forces  should  be  resolved  into  their  vertical  and  horizontal 
components.  The  components  are  then  used  as  vertical  and 
horizontal  forces. 

75.  The"  Ladder. — The  ladder  problem  offers  an  excellent 
opportunity  for  the  application  of  the  laws  of  non-concurrent 
forces.     AD   (Fig.  55)  is  a  uniform  ladder  10  ft.  long  and 
weighing   14  Ib.     The  rungs  are  spaced   1  ft.   apart.     The 
weight  (W)  of  the  ladder  is  considered  as  a  force  of  14  Ib. 
acting  downward  at  the  center  of  gravity.     A  load  (L)  of 
40.25  Ib.  is  suspended  3  ft.  from  D.     The  upper  end  of  the 
ladder  is  just  held  free  from  the  wall  by  a  spring  balance. 
Hence  there  is  no  friction  between  the  ladder  and  the  wall. 
The  lower  end  of  the  ladder  rests  on  a  roller  skate  which  in 
turn  rests  on  platform  scales.     It  is  kept  from  slipping  by 
means  of  a  spring  balance.     The  function  of  the  roller  skate 


EQUILIBRIUM  OF  NON-CONCURRENT  FORCES         63 


is  to  eliminate  friction  between  the  ladder  and  the  platform 
scales.  The  angle  between  the  ladder  and  the  wall  is  measured 
and  found  to  be  44°.  The  spring  balance  readings  are  recorded 
also  the  reading  of  the  platform  scales  (These  values  are  used 
only  for  checking  purposes).  The  balance  readings  are  18.2 
Ib.  each  and  the  platform  scales  reading  minus  the  weight  of 
the  roller  skate  is  54.25  Ib. 


Platform  Scales 
FIG.  55. — Ladder  mounted  for  experimental  purposes. 

It  is  now  desired  to  find  H\,  H»,  V,  the  ground  reaction  (G.R.)  and 
the  angle  (0)  of  the  ground  reaction.     It  is  evident  that: 

1.  #1    =    #2, 

2.  V  =  L  +  W, 

3.  W  X  BD  sin  44°  +  L  X  CD  sin  44°  =  #2  X  AD  cos  44°. 

As  both  HI  and  H2  are  unknown,  it  is  necessary  to  take  the  moments 
about  D  as  in  (3). 

14  X  5  X  .695  +  40.25  X  3  X  .695  =  H2  X  10  X  .719, 

H2  =  18.43  Ib.  =  #1. 

Since  the  observed  readings  HI  and  Hz  were  18.2  Ib.  each,  it  is  clear 
that  the  computed  and  observed  values  check  very  closely. 

V  =  40.25  +  14  =  54.25  lb.~ 

The  above  value  of  V  checks  exactly  with  the  observed  value  shown 
by  the  platform  scales. 

The  ground  reaction  (G.R.)  is  the  resultant  of  Hi  and  V. 

Hi2  +  F2,    G.R.  =  Vffii2  +  V2,    G.R.  =  V339.66  +  2943.06 
Ib. 


G.R.2  = 

57.29. 


64 


MECHANICS 


Tan  (0)  =  V/Hl  =  54.25/18.43  =  2.94.  By  referring  to  the  trigono- 
metric tables  in  the  appendix,  the  angle  in  the  tangent  column  correspond- 
ing nearest  to  the  decimal  2.94  is  found  to  be  71°. 

76.  The  Wall  Crane. — Figure  56  represents  a  laboratory 
model  of  a  simple  wall  crane.  The  member  AB  (considered 
weightless)  is  held  at  rest  by  the  tension  (T)  in  the  tie  DB, 
the  wall  reactions  (V  and  H)  and  the  load  (L).  The  com- 
pression in  A B  is  designated  by  the  letter  C.  Since  AB  is  in 

equilibrium  the  following  con- 
ditions prevail: 

1.  H  =  C, 

2.  H  =  T  cos  6, 

3.  V  +  T  sin  B  =  L, 

4.  T  X  7  sin  8  =  L  X  4. 
Assuming  L  and  B  to  be  given, 

T  is  found  from  equation  (4). 
Substituting  the  value  of  T  in 
equation  (2),  we  get  the  values 
of  H  and  C.     Continuing  the  substitution  in  equation  (3), 
the  value  of  the  vertical  hinge  reaction  (V)  is  found. 
Questions  and  Problems 

1.  What  is  meant  by  non-concurrent  forces? 

2.  Under  what  conditions  will  non-concurrent  forces  be  in  equilibrium? 

3.  A  horizontal  bar  AB  4  ft.  long  and  weighing  10  lb.,  is  hinged  to  the 
wall  at  A.     B  is  supported  by  a  tie  attached  to  the  wall  at  D,  making 
an  angle  of  40°  with  the  vertical.     If  100  lb.  is  suspended  1  ft.  from  B, 
find  (a)  the  tension  in  the  tie  DB;  (6)  the  compression  in  AB;  (c)  the 
vertical  and  horizontal  reactions  at  A  and  D. 

D 


FIG.  56. — Wall  crane. 


lo*    ^ow  so^ 

FIG.  57. 

4.  In  Fig.  57,  find  the  tension  in  DB;  the  compression  in  AB;  and  the 
vertical  and  horizontal  reactions  at  A  and  D. 


EQUILIBRIUM  OF  NON-CONCURRENT  FORCES         65 

6.  A  uniform  door  8  ft.  high  and  3.5  ft.  wide,  weighing  50  lb.,  is  sup- 
ported by  two  hinges  1  ft.  from  the  top  and  bottom  respectively. 
Assuming  the  total  weight  of  the  door  to  be  borne  by  the  lower  hinge, 
find  the  vertical  reaction  at  the  lower  hinge  and  the  horizontal  pull 
exerted  by  the  upper  hinge. 

6.  A  ladder  20  ft.  long  makes  an  angle  of  40°  with  a  smooth  wall. 
If  the  ladder  weighs  35  lb.  and  its  center  of  gravity  is  9  ft.  from  the  foot 
of  the  ladder,  find  (a)  the  horizontal  reaction  of  the  wall;  (6)  the  vertical 
reaction  at  the  foot  of  the  ladder;  (c)  the  horizontal  thrust  at  the  foot  of 
the  ladder;  (d)  the  ground  reaction;  (e)  the.  angle  of  the  ground  reaction. 

7.  Repeat  Problem  6,  assuming  a  friction  of  6  lb.  at  the  top  of  the 
ladder. 

8.  Repeat  Problem  7,  if  a  painter  weighing  175  lb.  is  standing  12  ft. 
up  the  ladder. 


CHAPTER  XII 


COMMERCIAL  AND  LABORATORY  STRUCTURES 

77.  Trusses. — A  truss  is  a  system  of  members  (beams,  bars, 
rods,  etc.),  joined  together  by  pins  or  rivets,  to  form  a  rigid 
framework.  While  trusses  may  be  of  various  shapes,  modern 
engineering  favors  those  of  the  triangular  type, — consisting 
either  of  a  single  triangle  or  a  collection  of  triangles.  Triangu- 
lar trusses  are  preferred  on  account  of  the  fact  that  a  triangle 
is  the  only  geometrical  figure  which  can  not  change  its  shape 


FIG.  58. — Simple  truss. 


bed 
FIG.  59. — Compound  truss. 


without  changing  the  length  of  its  sides.  Trusses  are  so 
designed  that  the  members  are  subject  principally  to 
tension  or  compression.  As  a  rule,  extraneous  loads  act  at  the 
joints.  The  material  is  so  proportioned  that  deformations 
are  practically  negligible.  Steady  loads,  as  the  weight  of  the 
material,  etc.,  are  called  dead  loads;  varying  loads,  due  to 
wind,  snow,  etc.,  are  called  live  loads. 

Figure  58  represents  a  simple  truss.  The  applied  load  (L) 
produces  tension  in  CB  and  compression  in  AB. 

Figure  59  represents  a  compound  deck  truss  supported  at 
A  and  E,  with  loads  (L)  applied  as  shown.  Tension  members 
are  shown  with  light  lines;  compression  jnembers  with  heavy 


COMMERCIAL  AND  LABORATORY  STRUCTURES        07 

lines.  AE  and  bd  are  known  as  the  upper  and  lower  chords 
respectively.  AB,  BC,  CD,  DE,  be  and  cd  are  called  panels. 
A,  B,  C,  D,  E,  b,  c  and  d  are  joints  or  apexes.  The  vertical 
distance  h  is  the  height  or  depth  of  the  truss.  The  members 
between  the  chords,  Ab,  Bb,  Be,  Cc,  DC,  Dd  and  Ed,  are  web 
members  or  braces  and  are  distinguished  as  verticals  and 
diagonals.  Web  members  in  compression  are  usually  called 
struts;  tension  members  are  usually  called  ties  or  stays. 

78.  Roof  Trusses.  —  Roof  trusses  are  rigid  structures  of 
metal  or  wood  or  a  combination  of  the  two,  designed  to  support 
roofs.  A  few  of  the  standard  types  will  be  touched  upon 
below. 

The  Fink  truss  is  commonly  used  for  spans  which  do  not 


(a)   Fink.  (b)   Pratt. 


(c)  Howe.  (d)  Triangular. 

FIG.  60. — Types  of  roof  trusses. 

exceed  100  ft.  Figure  60 (a)  shows  a  40  ft.  span.  The  Fink 
truss  is  very  economical  due  to  the  relative  shortness  of  its 
struts.  The  Pratt  truss  (Fig.  60(6))  is  very  popular  and  is 
often  used  instead  of  the  Fink  truss.  The  Howe  truss  (Fig. 
60 (c))  is  frequently  used,  especially  if  wood  is  to  enter  into  the 
construction.  The  rafters  and  diagonals  are  generally  of 
wood,  the  verticals  of  steel  and  the  lower  chords  either  of 
wood  or  steel.  The  triangular  truss  (Fig.  60 (c?))  is  often  used 
for  short  spans. 

79.  Bridge  Trusses. — Bridge  trusses  are  divided  into  two 
classes:  railroad  bridges  and  highway  bridges.  A  few  of  the 
more  common  types  will  be  discussed  below. 

Figure  61  (a)  represents  a  riveted  Pratt  railroad  truss  used 


68 


MECHANICS 


for  spans  up  to  160  ft.  The  Warren  riveted  truss  (Fig.  61(6)) 
is  used  for  spans  from  120  to  160  ft.  It  is  cheaper  than  the 
Pratt  truss  and  just  as  satisfactory. 

The  quadrangular  Warren  truss  (Fig.  62 (a))  is  commonly 


(a)  Pratt.  (&)  Warren. 

FIG.  61. — Types  of  railroad  bridges. 


(a)   Quadrangular  Warren.  (6)  Baltimore. 

FIG.  62. — Types  of  highway  bridges. 

used  for  spans  from  80  to  170  ft.     The  Baltimore  truss  (Fig. 
162(6))  is  used  for  long  span  bridges. 

80.  Laboratory  Study  of  a  Roof  Truss. — ABC  is  a  labora- 
tory model  of  a  simple  roof  truss  (Fig.  63).     The  members 


FIG.  63. — Roof  truss  mounted  for  laboratory  use. 

AB  and  CB  weigh  2.5  Ib.  each  and  are  hinged  at  B.  The 
suspended  weight  W  produces  compression  in  AB  and  CB. 
The  compression  is  balanced  by  the  vertical  and  horizontal 
reactions  at  A  and  C.  Inspection  of  the  apparatus  shows  that 


COMMERCIAL  AND  LABORATORY  STRUCTURES        69 

the  compression  in  A  B  is  the  same  as  the  compression  in  CB 
and  that  the  reactions  at  A  are  the  same  as  the  reactions  at  C. 
It  is  assumed  that  one  half  of  the  weight  of  each  member  acts 
vertically  downward  at  its  extremities.  Thus  the  total  force 
acting  downward  at  B  is  W  +  2.5  Ib.  Balance  1  gives  the 
total  downward  force  at  A  and  balance  2  the  horizontal  reac- 
tion at  A .  The  balances  are  used  only  as  checks. 

The  following  experiment  was  done  in  the  mechanics  labora- 


28.5* 
FIG.  64. — Force  diagram  for  Fig.  63. 


tory  of  the  Wm.  L.  Dickinson  High  School.  A  weight  of  26 
Ib.  was  suspended  from  B,  making  the  total  downward  force 
26  +  2.5  or  28.5  Ib.  The  apparatus  was  adjusted  until 
BI  was  vertical  and  B2  horizontal.  BI  read  15.75  Ib.  and 
Bz  read  15  Ib.  Angles  ABW  and  CBW  were  measured  and 
found  to  be  46°  each.  « 

In  order  to  find  the  compression  in  AB  and  BC  and  the 
vertical  and  horizontal  reactions  at  A  and  C,  a  force  diagram 
was  constructed  as  shown  in  Fig.  64.  Using  a  scale  of  1  in. 
=  15  Ib.,  a  vector  or  force  line  was  drawn  to  represent  the 
total  downward  force  of  28.5  Ib.  This  force  was  resolved  by 
the  parallelogram  method  into  two  components  at  angles  of 


70  MECHANICS 

46°  with  the  force.  Each  component  was  found  to  be  20.6 
Ib.  Thus  the  compression  in  AB  and  BC  is  20.6  Ib. 

It  is  evident  that  the  compression  in  the  member  AB  is 
just  equal  and  opposite  in  direction  to  the  resultant  of  the 
vertical  and  horizontal  reactions  V  and  H  at  A .  Accordingly, 
the  compressive  force  (20.6  Ib.)  was  resolved  into  its  vertical 
and  horizontal  components  by  the  parallelogram  method. 
It  should  be  noted  that  the  same  scale  is  used  throughout. 

V  was  found  to  be  14.5  Ib.  Since  one  half  of  the  weight  of 
the  member  AB  acts  at  A,  it  is  necessary  to  add  1.25  Ib.  to  V 
before  checking  with  BI.  Then  the  computed  value  of  V 
checks  exactly  with  the  actual  value  as  shown  by  the  spring 
balance.  H  was  found  to  be  15  Ib.,  checking  exactly  with  B2. 
As  previously  stated,  the  reactions  at  C  are  the  same  as  at  A 
and  the  compression  in  BC  is  the  same  as  in  A  B.  The  results 
of  the  experiment  were  tabulated  as  follows. 

Weight  of  member  AB 2 . 50  Ib. 

Weight  of  member  BC 2 . 50  Ib. 

Weight  suspended  (TF) 26.00  Ib. 

Total  downward  force  at  B 28. 50  Ib. 

Reading  of  balance  1 15 . 75  Ib. 

Reading  of  balance  2 15 . 00  Ib. 

Angle  ABW 46°. 

Angle  CBW. 46°. 

Calculated  vertical  component  of  AB 14 . 50  Ib. 

H  weight  of  AB  (acting  on  A) 1 . 25  Ib. 

Total  calculated  downward  force  at  A 15 . 75  Ib. 

(check)  Vertical  force  recorded  by  balance  1 15 . 75  Ib. 

Difference 0 

Calculated  horizontal  component  of  AB 15.00  Ib. 

(check)  Horizontal  force  recorded  by  balance  2 15 .00  Ib. 

Difference 0 

81.  Laboratory  Study  of  a  Simple  Truss. — Figures  65  and 
67  represent  a  laboratory  model  of  a  simple  truss  or,  as  it  is 
often  called,  a  "stick  and  tie."  BC  is  a  light  compression 
member  whose  weight  may  be  disregarded,  fitting  into  a  slot 
at  B.  The  tie  AC  is  a  tension  member.  Balance  1  reads  the 
tension  (T)  in  AC  and  balance  2  reads  the  compression  in 


COMMERICAL  AND  LABORATORY  STRUCTURES       71 

BC.  These  values  are  used  as  a  check.  Balance  1  is  read 
directly.  Balance  2  is  pulled  until  BC  just  begins  to  leave 
the  slot  at  B.  The  balance  reading  at  this  exact  instant 
measures  the  compression  in  BC. 

The  following  figures  were  obtained  by  two  students  in  the 
Wm.  L.  Dickinson  High  School. 

CASE  I.     BC  HORIZONTAL 

A  weight  of  10.75  Ib.  was  suspended  from  C  and  the  appara- 
tus adjusted  until  BC  was  level.     The  angle  ACS  was  meas- 


FIG.  65. — Simple  truss  mounted  for     FIG.  66. — Force  diagram  for  Fig.  65. 
laboratory  use. 

ured  and  found  to  be  35°.  The  balance  readings  were  taken 
and  recorded.  Balance  1  was  18.5  Ib.  and  balance  2  was 
15.25  Ib.  Using  C  as  a  free  body,  T  and  R  were  computed  by 
trigonometry.  Referring  to  Fig.  66,  it  is  clear,  since  point 
C  is  in  equilibrium,,  that : 

1.  T  cos  ACY  =  WorT  =  W/  cos  ACY, 

2.  R  =  T  cos  ACB. 

First  T  was  found  in  equation  (1)  and  then  its  value  was 
substituted  in  equation  (2)  to  find  R. 

1.  T  =  W/  cos  55°,  T  =  10.75/.574  =  18.70  Ib., 

2.  R  =  T  cos  35°,  ft  =  18,70  X  .819,  =  15,31  Ib. 


72 


MECHANICS 


TABULATION 

Angle  ACE 35°. 

Angle  ACY 55°. 

Weight  suspended  (W) 10.75  Ib. 

T  (calculated  value) 18. 70  Ib. 

T  (observed  value,  balance  1) 18 . 50  Ib. 

Difference 20  Ib. 

R  (calculated  value) 15.31  Ib. 

R  (observed  value,  balance  2) 15 . 25  Ib. 

Difference 06  Ib 

CASE  II.     BC  NOT  HORIZONTAL 

In  this  case,  a  suspended  weight  of  10  Ib.  was  used  and  the 
apparatus  adjusted  until  BC  was  at  an  angle  as  shown  in  Fig. 
67.  The  angles  BCW  and  ACB  were  measured  and  found 

r 


Balance  /. 


alance 2 
C      10.25* 


FIG.  67. — Simple  truss  mounted  for 
laboratory  use. 


r* 


Tco$ACD=RcosBCD 
Teas  ACY+Rcos  BCW-W 
FIG.  68. — Force  diagram  for 
Fig.  67. 


to  be  60°  and  59°  respectively.  The  balance  readings  were 
recorded  as  a  check.  Balance  1  read  10.3  Ib.;  balance  2  read 
10.25  Ib. 

Using  C  as  a  free  body,  T  and  R  were  computed  by  trigono- 
metry. Referring  to  the  force  diagram  (Fig.  68),  we  see, 
since  point  C  is  in  equilibrium,  that: 

1.  T  cos  ACD  =  R  cos  BCD, 

2.  T  cos  ACY  +  R  cos  BCW  =  W. 

Since  T  and  R  are  both  unknown,  it  is  necessary  to  solve 
for  them  by  the  method  of  simultaneous  equations.  Putting 
in  all  known  values,  we  have: 


COMMERCIAL  AND  LABORATORY  STRUCTURES        73 

1.  T  cos  29°  =  R  cos  30°,  T  X  .875  =  R  X  .866, 

2.  T    cos    61°  +  R    cos    60°  =  W,    T  X  .485  +  72  X  .500 
=  10. 

Simplifying  and  removing  decimals : 

1.  875  T  =  866  R, 

2.  485  T  +  500  R  =  10,000. 

Since  T  =  866  72/875,  then,  in  equation  (2) : 

485  X  866  72/875  +  500  72  =  10,000, 
72  =  10.32  Ib. 

Substituting  the  value  of  72  in  equation  (1) : 

875  T  =  866  X  10.32, 
T  =  10.25  Ib. 

TABULATION 

Angle  BCW 60° 

Angle  ACB 59° 

Angle  BCD 30° 

Angle  ACD 299 

Weight  suspended  (W) 10  Ib. 

T  (calculated  value) 10. 25  Ib. 

T  (observed  value,  balance  1) 10.30  Ib. 

Difference 05  Ib. 

R  (calculated  value) 10.32  Ib. 

R  (observed  value,  balance  2) 10 . 25  Ib. 

Difference 07  Ib. 

82.  Laboratory  Study  of  a  Simple  Hoisting  Crane. — Figure 
69  represents  a  laboratory  model  of  a  simple  hoisting  crane. 
The  angles  and  suspended  weight  (W)  may  be  varied  as 
desired.  Given  Wt  the  weight  of  the  boom  AB  and  the 
necessary  angles,  the  tensions  (Ti  and  T%)  may  be  computed, 
as  well  as  the  compression  (72)  in  the  boom.  The  problem 
is  similar  to  that  of  the  simple  truss  just  studied.  Since  AB 
is  fairly  heavy,  its  weight  may  not  be  neglected.  One  half 
the  weight  of  A B  must  be  added  to  W  for  the  total  downward 
force  (L). 

Referring  to  Fig.  70  (force  diagram  for  Fig.  69),  it  is  evident 
that  772  has  the  same  tension  as  W  (not  L) .  Hence  Tz  =  W. 


74 


MECHANICS 


It  is  also  evident  that : 

(a)  Ti  cos  35°  +  Tz  cos  53°  =  R  cos  65°,  and 
(6)  !Fi  cos  55°  +  R  cos  25°  =  T2  cos  37°  +  L. 
Since  Tz  =  W,  then  TI  and  72  are  the  only  unknowns.     T\ 
and  #  are  found  by  means  of  simultaneous  equations  as  illus- 
trated in  the  simple  truss,  previously  studied.     T\  is  checked 
by  the  balance  C  and  R  is  checked  by  pulling  out  on  AB  with  a 
spring  balance. 


FIG.  69. — Laboratory 
type  of  hoisting  crane. 


0      A  L 

FIG.  70. — Force  diagram  for  Fig.  69. 


83.  Laboratory  Study  of  Shear  Legs. — Figure  71  represents 
a  laboratory  model  of  a  pair  of  shear  legs  such  as  is  found 
around  docks,  etc.  It  will  be  seen  that  equilibrium  is  produced 
by  the  tension  in  OD  and  OS  and  by  the  compression  in  the 
legs  OA  and  OC.  It  is  desired  to  find  OD,  OA  and  OC.  The 
apparatus  is  set  up  as  shown  and  the  tension  in  OD  is  checked 
by  the  balance  B\.  The  compression  in  OA  is  checked  by 
pulling  on  B2  along  the  line  OA  until  A  just  begins  to  move 
from  its  support.  The  compression  in  OC  is  checked  simi- 


COMMERCIAL  AND  LABORATORY  STRUCTURES       75 

larly.     One  half  the  weight  of  each  leg  must  be  added  to  W 
for  the  total  load  (L)  at  0. 

Since  the  forces  are  not  all  in  the  same  plane,  assume  OA 
and  OC  to  be  replaced  by  a  single  force  OR  in  the  plane  DOW. 
It  is  evident  that  the  imaginary  force  OR  is  the  resultant  of 
OD  and  OS.  Given  the  total  downward  force  at  0  (L)  and  the 


w 


FIG.  71. — Laboratory  type  of  shear  legs. 


angles  a  and  6,  OD  and  OR  are  found  graphically  by  the 
parallelogram  method  or  trigonometrically. 

Inspection  of  Fig.  71  shows  that  OR  (just  computed)  is  the 
resultant  of  OA  and  OC.  Since  these  forces  are  in  the  same 
plane,  OR  is  resolved  either  graphically,  or  trigonometrically 
into  its  components  OA  and  OC:  Thus  the  compression  in 
the  legs  is  determined. 


76  MECHANICS 

Questions  and  Problems 

1.  What  is  meant  by  a  truss? 

2.  Why  is  a  triangular  truss  favored  by  engineers? 

3.  Draw  a  compound  deck  truss  and  label  each  part. 

4.  What  is  a  roof  truss?     Make  a  drawing  to  show  four  common  types. 

5.  What  is  a  bridge  truss ?     Into  two  what  classes  may  they  be  divided? 
Show,  by  drawings,  two  common  types  of  each. 

6.  Describe  a  laboratory  experiment  to  determine  the  static  forces  in 
(a)  a  simple  roof  truss;  (b)  a  "stick  and  tie;"  (c)  a  hoisting  crane;  (d)  a 
pair  of  shear  legs. 

7.  Find  the  compression  in  BA  and  BC  and  the  vertical  and  horizontal 
reactions  at  A  and  C  in  Fig.  63,  if  the  angles  remain  the  same  and  W 
equals  20  Ib. 

8.  Solve  for  T  and  R  in  Fig.  65,  if  the  angles  remain  the  same  and  W 
equals  12  Ib. 


CHAPTER  XIII 
ELASTICITY 

84.  Elasticity  Explained. — Whenever  a  body  is  acted  upon 
by  an  external  force,  a  change  in  the  shape  of  the  body  is 
produced.     The  applied  force  may  act  in  three  ways:  (1) 
it  may  stretch  the  body;  (2)  it  may  compress  the  body;  (3) 
it  may  shear  the  body.     If  the  body  has  not  been  loaded  too 
heavily,  it  will  return  to  its  original  shape  as  soon  as  the 
displacing  load  has  been  removed.     This  is  a  manifestation 
of  elasticity,  a  property  which  all  matter  possesses.     Steel, 
glass,  wood,  rubber,  etc.  are  elastic. 

The  amount  of  elasticity  is  measured,  not  by  the  ease  with 
which  the  distortion  is  effected,  but  by  the  difficulty  with  which 
the  distortion  is  effected.  Steel  is  exceedingly  elastic,  while 
rubber  is  only  moderately  so.  A  unit  force,  applied  to  a 
unit  length  of  steel,  will  produce  much  less  distortion  than  the 
same  force,  applied  to  a  piece  of  rubber  of  like  length  and 
cross-section.  Steel,  therefore,  is  more  elastic  than  rubber. 

85.  Elasticity   Defined. — Elasticity   is   a   general   property 
of  matter  in  consequence  of  which  all  bodies  having  undergone  a 
change  in  shape,  tend  to  resume  their  original  shape  as  soon  as 
the  displacing  force  has  been  removed. 

86.  Hooke's  Law. — //  the  applied  forces  are  not  too  large, 
elastic  deformations  of  all  kinds  are  directly  proportional  to 
the  forces  producing  them.     This  law  was  first  enunciated  and 
demonstrated  by  Robert  Hooke,1  an  Englishman,  and  bears 
his  name. 

Let  us  analyze  the  above  law  in  order  to  understand  clearly 
what  is  meant.  Suppose  that  we  have  a  piece  of  copper  wire 
suspended  from  a  rigid  support  and  that  the  free  end  bears 

1  Robert  Hooke  (1635-1703).  English  physicist.  Made  many 
inventions  in  physical  and  astronomical  instruments. 

77 


78 


MECHANICS 


•Rigid  Support 


l 


a  scale  pan  of  sufficient  weight  to  take  all  the  kinks  out  of  the 
wire.  Suppose,  further,  that  a  pointer  moving  over  a  gradu- 
ated scale  is  attached  a  slight  distance  above  the  scale  pan. 
If  a  weight  of  5  Ib.  is  placed  in  the  scale  pan  and  the  pointer 
moves  down  2  graduations,  then  a  weight  of  10  Ib.  will  cause 
it  to  move  down  4  graduations,  and  a  weight  of  15  Ib.  will 

cause  it  to  move  down  6 
graduations.  In  short,  if 
we  double  the  pull,  we 
double  the  stretch;  if  we 
triple  the  pull,  we  triple 
the  stretch.  In  each  case, 
the  pointer  will  return  to 
its  first  position  as  soon  as 
the  weight  is  removed. 

The  apparatus  shown  in 
Fig.  72  is  very  convenient 
for  studying  Hooke's  law. 
A  piece  of  german  silver 
wire  104.5  in.  long  and 
.0005  sq.  in.  in  cross-sec- 
tion is  firmly  clamped  at 
the  top  to  a  rigid  wall  sup- 
port  (A).  The  opposite 
end  of  the  wire  carries  a 
scale  pan  in  which  the 
various  loads  (F)  are 

FIG.  72.— Apparatus    to    verify    Hooke's    niflppj        TI         nnintpr    (T)\ 

law  for  tension.  placed.     I  ne   pointer  (L>) 

is  attached  to  the  wire  at 

B  an  dsispivoted  at  C.  It  will  be  seen  that  as  B  moves 
down,  D  will  move  up.  In  order  to  magnify  the  elongations 
of  the  wire  and  make  more  accurate  reading  possible,  DC 
is  made  10  times  as  long  as  BC.  Hence,  the  movement  of 
the  pointer  along  the  scale  is  always  10  times  as  much  as 
the  actual  elongation  of  the  wire. 

The  following  experiment  was  performed  by  a  student  in 
the  Wm.  L.  Dickinson  High  School  with  the  apparatus  and 


f  Pivot        (Pointer 


Scale  Pat 


ELASTICITY 


79 


material  just  described.  First  a  straightening  load  of  10  Ib. 
was  placed  on  the  scale  pan  to  take  the  kinks  out  of  the  wire. 
This  weight  is  not  counted  in  the  calculations  and  is  called 
the  "zero"  load.  The  pointer  (D)  was  adjusted  so  that  it 
was  somewhat  below  the  center  of  the  scale  (E)  and  the  scale 
was  moved  until  the  pointer  just  coincided  with  an  even  scale 
division.  A  weight  of  5  Ib.  was  added  to  the  scale  pan  and 
left  2  min.  On  being  removed,  it  was  found  that  the  pointer 
returned  exactly  to  its  original  position,  showing  the  apparatus 
to  be  properly  adjusted.  If  the  pointer  had  not  returned  to 
its  original  position,  there  was  either  a  loose  connection  or  a 
sagging  of  the  wall  support. 

Again  the  pointer  reading  at  "zero"  load  was  taken  and 
recorded.  A  load  of  2  Ib.  was  now  placed  in  the  scale  pan 
and,  after  2  min.,  the  pointer  reading  was  carefully  taken. 
The  loads  were  increased,  2  Ib.  at  a  time,  until  8  Ib.  had  been 
added,  the  pointer  reading  being  recorded  in  each  case.  The 
results  were  tabulated  as  follows: 


Loads  in  Ib. 

Pointer  readings 
in  in. 

Apparent  elon- 
gations in  in. 

Actual  elon- 
gations in  in. 

0 

2.75 

2 

3.00 

0.25 

0.025 

4 

3.25 

0.50 

0.050 

6 

3.50 

0.75 

0.075 

8 

3.75 

1.00 

0.100 

A  graph  was  made  from  the  above  data  and  is  shown  on 
page  80.  Note  that  the  load-elongation  line  is  straight,  denot- 
ing a  direct  proportion. 

87.  The  Elastic  Limit. — If  the  german  silver  wire,  used  in 
the  experiment  just  described,  had  been  subjected  to  further 
load,  a  point  would  have  been  reached  at  which  the  loads  and 
deflections  would  have  not  been  in  direct  proportion.  The 
stretches  would  then  increase  in  greater  proportion  than  the 
loads,  and  the  load-elongation  line  would  bend  to  the  right. 


80 


MECHANICS 


The  pointer  would  not  return  to  its  original  position  when  the 
loads  were  removed,  and  eventually  the  wire  would  break. 

The  stress  beyond  which  an  elastic  material  will  not  return  to 
its  original  position  when  the  load  is  removed,  is  called  the  elastic 
limit  for  that  material.  The  elastic  limit  is  difficult  to  deter- 
mine exactly,  but  may  be  determined  very  closely. 

88.  The  Yield  Point. — After  a  material  passes  its  elastic 
limit,  a  stress  is  reached  at  which  the  elongations  continue 
without  the  addition  of  further  load.  This  is  called  the  yield 
point  and  shows  that  rupture  is  about  to  take  place. 


fl 

y 

x 

^ 

. 

X 

0 

. 

:V 

~s~ 

Q_ 

c  4 

&$ 

^ 

D 

\( 

^ 

3 

••) 

„/ 

/ 

c 

X 

~^~ 

n 

g 

/ 

3 

.0! 

50 

.1C 

0 

Stretches  in  Inches 
FIG.  73. — Graph  showing  relation  of  load  and  stretch  for  german  silver  wire. 

89.  Stress. — Whenever  an  elastic  body  is  acted  upon  by 
a  force,  it  is  said  to  be  under  stress.  The  stress  may  be  the 
result  of  a  tensile  load,  in  which  the  molecules  tend  to  separate ; 
it  may  be  the  result  of  a  compressive  load,  in  which  the  mole- 
cules tend  to  crowd  together;  or  it  may  be  the  result  of  a 
shearing  load,  in  which  certain  particles  are  caused  to  slide 
past  others.  In  every  case,  the  body  resists  the  effort  to 
change  its  molecular  arrangement. 

The  internal  resistance  of  a  body  to  a  change  of  shape  is  called 
stress.  It  is  determined  in  amount  by  dividing  the  external 
force  acting  in  Ib.  by  the  area  in  sq.  in.  over  which  the  force 
acts.  The  area  must  always  be  taken  at  right  angles  to  the 
force.  Let  us  take  a  practical  illustration. 


ELASTICITY  81 

Suppose  a  steel  wire  .10  sq.  in.  in  cross-section  is  bearing  a 
suspended  load  of  50  Ib.  The  load  of  50  Ib.  is  evenly  dis- 
tributed over  the  area  of  the  wire.  If  we  wish  to  use  a  wire 
of  1  sq.  in.  cross-section  and  keep  the  same  rate  of  tension,  we 
must  use  a  load  of  500  Ib.  In  either  case,  the  wire  is  under  a 
stress  of  500  Ib.  per  sq.  in.  The  stress  is  simply  the  load  applied 
in  Ib.  per  sq.  in.1 

external  force  acting  in  Ib. 
Stress  = 


area  in  sq.  in.  over  which  the  force  acts 

90.  Strain. — Whenever  a  body  is  subjected  to  a  stress,  a 
change  of  shape  results.     The  amount  of  change  of  size  per 
unit  of  original  size  is  called  strain.     To  illustrate :  suppose  we 
have  a  piece  of  copper  wire  about  .04  in.  in  diameter  and  72 
in.  long.     Under  a  load  of  5  Ib.,  the  wire  will  elongate  .019  in. 
If  a  load  of  10  Ib.  is  used,  the  wire  will  elongate  .038  in.     In 
the   first    case,    the  strain  will  be  .019/72  or   .000263  inch 
per  inch. '     In  the  second  case,  it  will  be  .038/72  or  .000526 
inch  per  inch.2 

0 .     .         change  of  size  in  inches 
Strain  =  — r-? — y—. — = — : — =— 
original  size  in  inches 

91.  Young's   Modulus. — The   mathematical   relation   of  the 
stress  and  the  strain  is  called  Young's  modulus. 

TZ         »         j  -i  stress 

Young  s  modulus  = — 

strain 

Young's  modulus  is  also  called  the  coefficient  or  modulus  of 
elasticity.  It  is  practically  the  same  for  tension  as  compres- 
sion. The  modulus  for  shearing  strains  will  be  considered 
later.  The  modulus  for  compression  and  tension,  in  the  case 
of  steel,  will  average  about  30,000,000  Ib.  per  sq.  in.;  for 
copper  about  15,000,000  Ib.  per  sq.  in.  These  figures  will 
vary  somewhat,  but  maybe  taken  as  very  close  approximations. 

We  have  seen,  from  the  experiment  with  the  german  silver 
wire,  that  elastic  deformations  are  directly  proportional  to  the 
forces  producing  them,  provided  the  elastic  limit  is  not  ex- 

1  May  also  be  expressed  in  Kg.  /cm2. 

2  May  also  be  expressed  in  cm.  /cm. 


82 


MECHANICS 


ceeded.  This  was  stated  as  Hooke's  law.  We  may  now  state 
Hooke's  law  thus :  Within  the  elastic  limit,  the  strain  is  directly 
proportional  to  the  stress.  Figure  74  shows  this  relation  graph- 
ically. The  data  was  taken  from  the  experiment  with  the 
german  silver  wire  and  the  stress  and  strain  for  each  load  was 
computed.  The  following  graph  should  be  carefully  studied 
and  thoroughly  understood  by  the  pupil. 


16000 


1 


8000 


0  .00050  '.00100 

Inches  per  Inch 

FIG.  74. — Graph  showing  relation  of  stress  and  strain  for  german  silver  wire. 

92.  Ultimate  Strength. — The  ultimate  strength  of  an  elastic 
material  is  the  load  in  Ib.  per  sq.  in.  necessary  to  break  or  rupture 
the  material.1    The  rupture  may  be  due  to  a  tensile  stress,  a 
compressive  stress  or  a  shearing  stress.     Suppose  a  steel  bar 
.5  sq.  in.  in  cross-section  ruptures  under  a  load  of  50,000  Ib. 
The  ultimate  strength  is  50,000/.5  or  100,000  Ib.  per  sq.  in. 
The  ultimate  strength  for  a  body  under  tension  is  often  called 
tensile  strength.     The  tensile  strength  of  mild  steel  will  average 
about  70,000  Ib.  per  sq.  in.     It  will  vary  somewhat  due  to 
composition,  temperature,  etc. 

93.  Factor  of  Safety. — The  ratio  of  the  breaking  load  to 
the  working  load  for  any  material,  is  called  the  factor  of  safety 
for  that  material.     The  factor  of  safety  will  necessarily  depend 
on  the  nature  of  the  load.     Under  a  steady  load,  steel  should 
have  a  factor  of  4;  under  a  varying  load  a  factor  of  6;  under  a 

1  May  also  be  expressed  in  Kg. /era.2. 


ELASTICITY 


83 


load  in  which  the  stresses  are  liable  to  occur  sharply,  a  factor 
of  10.  In  machines  subject  to  sudden  overloads,  the  factor 
of  safety  is  made  very  high  to  avoid  any  chance  of  fatal  acci- 
dents. It  is  evident  that  all  members  which  have  to  bear 
frequent  and  sudden  overloads  should  have  a  high  percentage 
of  overs trength. 

94.  Elastic  Fatigue. — Practical  experience  and  experiment 
teaches  us  that  an  elastic  material  is  more  liable  to  break 
under  a  repeated  load  than  a  steady  load  of  the  same  size. 
This  is  especially  true  when  the  stresses  consist  both  of  ten- 
sions  and   compressions.     The   tendency  to  rupture   under 
repeated  strain  is  due  to  what  is  known  as  elastic  fatigue  and 
results  in  the  destruction  of  the  cohesive  force  at  the  point  of 
rupture.     The  rear  axle  of  an  automobile  occasionally  breaks. 
This  is  generally  due  to  repeated  twisting  strains.     A  piece 
of  copper  wire  twisted  back  and  forth  will  soon  undergo 
elastic  fatigue  and  break  at  the  point  of  strain. 

95.  Shear  Explained. — When  a  body  is  subjected  to   a 


J  ~, 


'£' 

1 

I 

H 

/J7" 

FIG.  75. — Diagram  to  illustrate  simple  shear. 

shearing  stress,  certain  particles,  of  which  the  body  is  com- 
posed, tend  to  slide  by  other  particles.  This  is  well  illus- 
trated in  the  case  of  a  pair  of  metal  shears,  the  metal  being 
sheared  apart  where  the  blades  come  together.  Planing, 
turning  and  punching  are  shearing  actions. 

In  order  to  understand  the  nature  of  a  simple  shear,  let 
us  refer  to  Fig.  75.  Here  we  have  a  rectangular,  elastic  body 
ABC  4  X  4  in.  on  the  top  face  and  2  in.  high.  Assume  the 


84  MECHANICS 

face  ABC  to  be  attached  rigidly  to  the  surface  on  which  it 
rests,  so  that  there  can  be  no  movement  at  the  bottom.  A 
force  of  80  Ib.  is  attached  to  the  face  EFGH  and  is  evenly 
distributed  over  the  16  sq.  in.  This  force  (L)  will  move  the 
horizontal  layers  to  the  right,  the  movement  varying  from 
maximum  at  the  top  to  zero  at  the  bottom.  Each  succeeding 
upper  layer  will  have  a  greater  displacement  to  the  right. 
Actually,  there  is  also  a  bending  tendency  present.  It  is 
sufficient  for  our  purpose,  however,  to  consider  only  the  simple 
shear,  leaving  the  rest  to  more  advanced  books. 

96.  Shearing  Stress. — Shearing  stress  is  equal  to  the  force 
acting  in  Ib.  divided  by  the  area  in  sq.  in.  over  which  the  force 
acts.     In  Fig.  75,  the  shearing  stress  is  80/16  or  5  lb./in.2 
It  may  also  be  expressed  in  Kg./cm.2 

97.  Shearing    Strain. — Shearing    strain    is    equal    to    the 
horizontal  movement  in  inches  of  a  particle  of  a  body  1  inch  above 
the  plane  to  which  it  is  attached.     Referring  to  Fig.  75,  let 
AJ  =  1  inch  and  JJ'  =  .0006  inch.     The  shearing  strain  is 
therefore    .0006    inch   per   inch.     Suppose   that  EE'  =  .001 
inch  and  that  AE  =  3  inches.     The  shearing  strain  will  then 
be  .001/3  or  .00033  inch  per  inch.     Shearing  strain  may  also 
be  expressed  in  cm./cm. 

98.  Modulus  of  Rigidity. — The  modulus  of  rigidity  (also 
called  the  modulus  of  elasticity  for  shear)  is  the  ratio  of  the 
shearing  stress  and  shearing  strain. 

,,   ,  ,       t    ,  .,..         shearing  stress 

Modulus  of  rigidity  = -= r-^      — —  • 

shearing  strain 

99.  Practical  Illustrations  of  Shear. — The  rivets  used  in 
fastening  together  boiler  plates,  etc.,  are  subject  to  shearing 
stresses.     Figure  76  shows  a  single-riveted  lap  joint.     The 
rivet  is  subject  to  a  shearing  stress  at  "a"  and  will  fail  at  that 
point  if  the  stress  is  of  sufficient  magnitude.     This  is  an 
example  of  single  shear. 

Figures   77   and   78   represent   single   and   double-riveted 
butt  joints.     The  rivets  are  liable  to  failure  both  at  "6"  and 


ELASTICITY 


85 


"c"  and  are  said  to  be  in  double  shear.     Figure  79  shows 
rivets  having  failed  under  single  and  double  shear  respectively. 

Illustrative  Problem. — What  force  is  required  to  punch  a  %  in.  hole  in  a 
steel  plate  %  in.  thick?  Shearing  strength  of  the  material  is  40,000 
lb./in.2 

Solution.— Area  under  shear  =  .75  X  3.1416  X  .50  =  1.178  sq.  in. 
1.178  X  40,000  =  47,120  Ib.  Ans. 

Illustrative  Problem. — The  head  of  a  1  in.  machine  bolt  is  %  in.  thick. 
Find  (1)  the  force  necessary  to  shear  the  bolt  off  at  right  angles  to  its 


FIG.  76. — Single-riveted 
lap  joint. 


FIG.  77. — Single-riveted 
butt  joint. 


Y///////// 


\S\N1 IfSxy 


D 

a 


FIG.  78. — Double-riveted  butt  joint. 


FIG.  79. — Failure 
of  rivets  due  to  single 
and  double  shear. 


length;  (2)  the  shearing  stress  tending  to  strip  the  head  from  the  bolt 
due  to  an  applied  force  of  10,000  Ib.  Shearing  strength  of  the  material 
is  40,000  lb./in.2 

Solution.— (1)  Area  under  shear  =  1  X  1  X  .7854  =  .7854  sq.  in. 
.7854  X  40,000  =  31,416  Ib.  Ans. 

2.  Area  under  shear  =  1  X  3.1416  X  .75  =  2.356  sq.  in.  10,000/ 
2.356  =  4,244  lb./in.2  Ans. 

Questions  and  Problems 

1.  Define  elasticity. 

2.  What  effects  may  be  produced  upon  an  elastic  body  by  an  external 
force? 

3.  State  and  explain  Hooke's  law. 

4.  Define  stress,  strain,  elastic  limit,  yield  point. 
6.  What  is  meant  by  Young's  modulus? 

6.  Describe  an  experiment  to  determine  the  relation  between  stress 
and  strain  for  an  elastic  body. 


86  MECHANICS 

7.  Define  ultimate  strength,  tensile  strength,  compressive  strength, 
shearing  strength,  factor  of  safety,  elastic  fatigue. 

8.  Define  shear  and  give  examples  of  shearing  actions. 

9.  Show,  using  diagram,  how  to  determine  stress  and  strain  for  simple 
shear. 

10.  What  is  meant  by  the  modulus  of  rigidity? 

11.  A  steel  strut  having  a  cross-section  of  4  sq.  in.  is  subjected  to  a 
compressive  force  of  2  tons.     Compute  the  stress. 

12.  A  uniform  steel  bar  2  sq.  in.  in  cross-section  ruptures  under  an 
applied  tensile  force  of  200,000  Ib.     What  is  the  ultimate  strength? 

13.  If  the  above  bar  is  of  wrought  iron,  how  large  a  force  will  be 
required  for  rupture?    Tensile  strength  of  wrought  iron  =  50,000  Ib./ 
in.2 

14.  A  wire  2  mm.  in  diameter  breaks  under  a  load  of  100  Kg.     If  the 
diameter  is  increased  to  4  mm.,  what  load  will  be  necessary  for  rupture? 

16.  A  steel  rod  6  ft.  long  and  1  sq.  in.  in  cross-section  is  subjected  to  a 
tensile  load  of  10,000  Ib.  Find  the  elongation  in  inches.  Modulus  of 
elasticity  =  30,000,000  lb./in.2 

16.  A  hard,  drawn  copper  rod  10  ft.  long  and  1  sq.  in.  in  cross-section 
stretches    .100   in.     Find    the  tensile  load  necessary  to   produce  the 
elongation.     Modulus  of  elasticity  17,600,000  lb./in.2 

17.  A  steel  bar  1  sq.  in.  in  cross-section  and  6  ft.  long  stretches  .024 
in.  under  an  applied  load  of  5  tons.     Determine  the  modulus  of  elasticity 
for  the  material. 

18.  A  horizontal  beam  4  sq.  in.  in  cross-section  is  securely  fastened  at 
one  end.     If  a  load  of  10,000  Ib.  at  right  angles  to  the  beam  is  applied 
10  in.  from  the  attached  end,  find  the  deflection  of  the  beam  at  the 
point  where  the  load  is  applied.     Modulus  of  rigidity  is  12,000,000  lb./in.2 

19.  A  steel  plate  %  in.  thick  has  an  ultimate  shearing  strength  of  40,000 
lb./in.2    What  force  is  required  to  punch  a  ^  in.  hole  through  the  plate? 

20.  Repeat  problem  19  for  a  %  in.  hole. 

21.  A  wrought  iron  rivet  Y±  in.  in  diameter  is  sheared  off  at  right 
angles  to  its  length.     If  the  shearing  strength  of  wrought  iron  is  40,000 
lb./in.2,  what  was  the  applied  force? 

22.  The  head  of  a  1  in.  steel  machine  bolt  is  %  in.  thick.     A  force  of 
12,000  Ib.  is  applied  in  such  a  way  that  there  is  a  tendency  to  strip  the 
head  from  the  bolt.     Find  the  shearing  stress  due  to  the  applied  load. 

23.  In  problem  22,  if  the  force  is  increased  gradually,  will  the  bolt 
fail  due  to  a  tensile  stress  or  a  shearing  stress?     Explain. 


CHAPTER  XIV 
WORK 

100.  Work  Defined   and   Explained. — Work  is  the  result 
accomplished  when  a  force  acts  through  a  distance.    The  force 
may  keep  the  motion  of  a  body  constant;  increase,  decrease  or 
change  the  direction  of  the  motion;  or  cause  the  body  to 
undergo  a  change  of  shape  or  size.     Common  examples  of 
work  are  a  hoisting  engine  lifting  concrete;  a  man  climbing  a 
mountain;  a  horse  drawing  a  wagon;  punching  holes  in  sheet 
iron;  stopping  an  automobile  by  means  of  the  brakes,  etc. 
Various    other   examples   will   occur   to    the    student.     The 
foundation  of  a  building,   although  it  exerts  an  enormous 
upward  force,   does  no  work,  because  the  weight  supported 
by  the  force  is  not  moved.     Force,  then  does  not  always  result 
in  work.     This  point  should  always  be  kept  in  mind. 

101.  How  Work  is  Measured. — The  work  done  upon  any 
body  is  measured  by  the  product  of  the  force  acting  and  the 
distance  through  which  the  force  acts. 

Work  done  =  Force  X  Distance  =  F  X  S 

102.  Units  of  Work. — Various  units  may  be  used  to  express 
the  amount  of  work  done.     If  a  10  Ib.  weight  is  lifted  verti- 
cally 10  ft.,  100  ft.-lb.  of  work  results.     The  same  work  may 
be  expressed  as  1,200  in.-lb.  or  .05  of  a  ft. -ton.     Further,  a 
weight  of  1  Kg.  lifted  vertically  through  a  height  of  10  meters 
results  in  10  kilogram-meters  of  work.     10  Kg.-m.  may  also 
be  expressed  as  10,000  g.-m.  or  1,000,000  g.-cm.     The  more 
common  units  are  given  below. 

UNITS  OF  WORK 

(a)  The  erg  is  the  work  done  when  a  force  of  1  dyne1  acts 

1  1  dyne  =  .00102  g.     1  Ib.  =  445,000  dynes. 

87 


88 


MECHANICS 


through  a  displacement  of  1  centimeter.  Since  the  erg  is  a 
very  small  unit,  the  joule  is  often  used.  A  joule  is  equal  to 
10,000,000  ergs. 

(6)  The  kilogram-meter  is  the  work  done  when  a  force  of  1 
kilogram  acts  through  a  displacement  of  1  meter. 

(c)  The  foot-pound  is  the  work  done  when  a  force  of  1  pound 
acts  through  a  displacement  of  1  foot. 

103.  Time  Not  a  Factor  in  Work. — The  element  of  time  is 
not  considered  in  an  expression  of  work.     In  a  scientific  sense, 
work  refers  only  to  the  result  accomplished,  regardless  of  the 
time  consumed  in  obtaining  the  result.     For  example,  it  will 
require  20,000  ft.-lb.  of  work  to  raise  1,000  Ib.  of  pig  iron 
through  a  vertical  height  of  20  ft.     It  is  evident  that  the 
amount  of  work  will  be  the  same  if  the  iron  is  lifted  all  in  one 
operation  or  if  each  pig  is  lifted  separately. 

104.  Further  Discussion  of  Work. — If  a  body  weighing 
100  Ib.  is  raised  vertically   10  ft.,   1,000  ft.-lb.  of  work  is 


Disp/acemeni-  =/o' 


SO'1 


30( 


EFFECTIVE  FORCE= 
50 cos 30°=  43.3 Lb.    ^  . 

FIG.  80. — Work  done  equals  433  ft.-lb. 


done.  If  the  same  body 
is  pulled  10  ft.  horizontally 
across  a  floor,  less  work 
is  done  as  the  force  re- 
quired to  pull  is  less  than 
the  force  required  to  lift. 
The  pulling  force  will 
depend  on  the  friction 
between  the  bodies  in  contact,  and  is  always  less  than  the 
lifting  force. 

In  case  it  is  desired  to  com- 
pute the  work  done  by  a  force 
acting  at  an  oblique  angle,  it 
is  necessary  to  find  the  com- 
ponent of  the  force  in  the 
direction  in  which  the  body 
moves.  Suppose  (Fig.  80) 
it  is  desired  to  find  the  work 

done  by  a  force  of  50   Ib.,  acting  at  an  angle  of  30°.     The 
body  is  moved  horizontally  10  ft,    The  work  done  is  not 


FIG.  81. 


WORK  89 

500  ft.-lb.     It  is  433  ft.-lb.  and  is  computed  as  follows: 

The  horizontal  component  of  50  Ib.  =  50  X  cos  30°  =  50  X  .866  = 
43.3  Ib.  43.3  X  10  =  433  ft.-lb. 

The  work  done  in  moving  a  body  uniformly  up  a  frictionless 
inclined  plane  may  be  determined  in  two  ways.  Suppose, 
referring  to  Fig.  81,  that  a  force  of  5  Ib.  is  necessary  to  move  a 
body  weighing  10  Ib.  up  a  plane  20  ft.  long  and  10  ft.  high. 

The  work  =  weight  X  height  =  10  X  10  =  100  ft.-lb.  The  work 
also  =  force  X  length  =  5  X  20  =  100  ft.-lb. 

105.  Work  Diagrams. — It  is  often  convenient  to  represent 


1  Space  *lfb 

FIG.  82. — Diagram  representing  40  ft.-lb.  of  work. 

work  done  by  a  work  diagram, — in  which  the  amount  of  work 
is  proportional  to  the  area  of  the  diagram.  Suppose  we  wish 
to  represent  a  constant  force  of  8  Ib.  acting  through  a  distance 
of  5  ft.  Adopting  some  convenient  scale  (such  as  1  space  = 
2  Ib.  and  1  space  =  1  ft.),  we  construct  a  diagram  as  shown 
in  Fig.  82.  It  is  evident  that  each  square  space  =  2  ft.-lb. 
and  that  the  total  number  of  square  spaces  represent  the  entire 
work  done  or  40  ft.-lb.  (20  X  2). 

Figure  83  represents  the  net  work  done  by  the  head  end  of  a 
Corliss  steam  engine  cylinder  on  1  sq.  in.  of  the  piston  during 
one  revolution  of  the  crank  shaft.  The  diagram  was  taken 
by  means  of  a  device  called  a  steam  indicator.  The  line 
abcdea  represents  the  steam  pressure  at  any  point  in  the  out- 


90  MECHANICS 

ward  and  return  stroke.  It  is  evident  that  the  effective 
pressure  (P)  at  any  chosen  place  will  be  proportional  to  a 
vertical  line  (ordinate)  connecting  the  upper  and  lower 
extremities.  By  drawing  a  series  of  such  ordinates  at  equal 
intervals  and  averaging  them,  the  mean  effective  pressure  for 
the  entire  working  stroke  can  be  determined,  if  the  vertical 
scale  is  known.  Since  1  in.  =  70  lb.,  the  mean  effective 


Stroke  =2  f+. 


FIG.  83.  —  Diagram  representing  work  done  by  the  crank  end  of  a  Corliss 
steam  engine  cylinder  during  one  revolution  of  the  crank. 

pressure  will'  be  about  35  Ib./sq.  in.  for  the  diagram  shown 
here.  The  stroke  is  2  ft.;  hence  the  useful  work  on  one  sq.  in. 
of  the  head  end  of  the  piston  for  one  revolution  will  be  70  ft.- 
Ib.  (35  X  2).  70  ft.-lb.  also  represents  the  area  (average 
height  X  length)  of  the  diagram  in  terms  of  the  scale  used. 

The  student  should  note  that  :  a  =  steam  admitted  into  the 
cylinder;  b  =  steam  is  cut  off  ;  c  =  exhaust  opens;  d  =  exhaust 
stroke  begins;  and  e  =  exhaust  closes  and  compression  begins. 

Questions  and  Problems 

1.  What  is  meant  by  work?     Give  various  examples  of  work. 

2.  What  is  the  general  formula  for  work  done? 

3.  Name  and  define  the  units  of  work. 

4.  Discuss  "time  as  an  element  in  work." 

5.  State  two  ways  of  computing  the  work  done  on  an  inclined  plane. 

6.  A  force  of  1  dyne  acts  through  10  cm.     How  much  work  is  done? 

7.  An  automobile  weighing  2,000  Kg.  travels  1  Km.  up  a  2  per  cent. 
grade.     Find  the  work  done  in  Kg.-meters. 


WORK  91 

8.  An  elevator  weighs  1,500  Ib.  more  than  its  counter-weight.     How 
much  work  is  done  if  it  rises  50  ft.  ? 

9.  A  block  of  wood  weighing  4,200  Ib.  is  8  ft.  long,  4  ft.  wide  and  3 
ft.  thick.     If  the  block  is  lying  on  the  8  ft.  X  4  ft.  side,  how  much  work 
is  necessary  to  turn  the  block  on  the  3  X  4  ft.  side? 

10.  How  many  foot-pounds  of  work  will  be  done  by  a  gasoline  engine 
of  15  per  cent,  efficiency  on  10  gallons  of  gas?     Assume  20,000  British 
thermal  units  per  Ib.  of  gas  (1  B.t.u.  =  780  foot-pounds). 

11.  A  hoisting  engine  burns  1  ton  of  coal  during  the  day.     If  the  engine 
is  2  per  cent,  efficient,  how  much  work  will  it  do  in  a  day?     (1  Ib.  coal  = 
13,000  B.t.u.) 

12.  A  force  acts  through  10  ft.  against  a  varying  resistance.     The 
successive  resistances  in  pounds  at  the  beginning  of  each  foot  are  as 
follows:  50,  55,  60,  58,  51,  52,  63,  60,  57,  49,  51.     Construct  a  work 
diagram,  drawn  to  scale,  and  determine  the  amount  of  work  done. 


CHAPTER  XV 
POWER 

106.  Power  Defined. — We  have  seen  that  time  is  not  an 
element  in  work.     For  example,  it  is  possible,  under  certain 
conditions,  for  a  light  horse  to  do  as  much  work  as  a  heavy 
horse,  provided  the  light  horse  is  allowed  more  time  to  com- 
plete his  task.     The  heavy  horse  is  said  to  possess  the  greater 
power,  however,  as  he  can  do  the  work  in  a  shorter  time. 
Power  concerns  itself  both  with  the  work  done  and  the  time. 
A  locomotive  will  do  many  times  as  much  work  in  a  given 
time  as  a  hoisting  engine  and  therefore  has  the  greater  power. 
Power  is  an  expression  of  the  rate  at  which  work  is  done. 

n  work  done 

Power  =  -  — — 
time 

107.  Units  of  Power. — The  units  of  power  in  general  use 
are  the  horsepower,  the  watt  and  the  kilowatt.     The  watt  and 
kilowatt  are  coming  into  popularity  very  rapidly.     Electric 
light  bulbs  are  rated  entirely  in  watts.     Electric  motors  and 
marine  engines  are  commonly  rated  in  kilowatts.     The  horse- 
power is  still  widely  used  and  is  very  important. 

108.  The   Horsepower. — James   Watt1   estimated   that   a 
heavy  work  horse,  traveling  at  an  average  rate  of  2.5  miles  per 
hour,  could  lift  a  weight  of  150  Ib.  by  means  of  a  rope  and 
pulley.     This  is  equivalent  to  33,000  ft.-lb.  of  work  per  minute 
or  550  ft.-lb.  per  second.     Watt's  estimate  is  probably  inaccu- 
rate, yet  it  furnishes  a  very  satisfactory  basis  for  comparisons 
of  power.     Whenever  work  is  being  done  at  the  rate  of  33,000 
ft.-lb.  per  minute,  one  horsepower  is  being  expended. 

ft.-lh.  of  work  per  minute 
Horsepower  -  33,000 

1  James  Watt  (1736-1819).  Scottish  engineer.  Invented  the  first 
condensing  steam  engine  and  the  first  centrifugal  governor.  First  used 
the  steam  indicator  to  determine  the  amount  of  work  done  by  a  steam 
engine. 

92 


POWER 


93 


109.  The  Watt  and  Kilowatt. — The  power  unit  in  the  c.g.s. 
system  is  an  erg  per  second.  For  practical  purposes,  it  is 
customary  to  use  a  joule  per  second  (10,000,000  ergs  per 
second).  The  latter  unit  is  called  a  watt  in  honor  of  James 
Watt.  One  horsepower  is  equivalent  to  746  watts.  Since 
there  are  1,000  watts  in  a  kilowatt,  it  is  evident  that  one 
horsepower  is  nearly  equivalent  to  %  of  a  kilowatt  and,  con- 
versely, that  one  kilowatt  is  slightly  more  than  %  of  a  horse- 
power. Electrically,  the  watt  is  the  work  done  in  one  second 
by  a  current  of  one  ampere  flowing  under  a  pressure  of  one 
volt. 

1  horsepower  (hp.)  =  746  watts 
1  kilowatt  (kw.)       =  1,000  watts 
1  horsepower  =  %  kw.  (approx.) 

1  kilowatt  =  l;H}hp.  (approx.) 


FIG.  84. — Prony  brake  mounted  for  power  test  of  an  electric  motor. 

110.  Brake  Horsepower  (b.hp.). — The  brake  method  is 
used  in  determining  the  power  delivered  by  a  rotating  shaft  or 
pulley.  It  is  also  known  as  the  absorption  method,  since  the 
power  transmitted  is  absorbed  and  the  mechanical  energy 
of  the  rotating  body  is  transformed  into  heat  energy.  A  device 
called  the  prony  brake  is  used  to  determine  brake  horsepower. 
It  is  described  in  the  succeeding  paragraphs. 

The  following  experiment  was  performed  by  a  student  in  the 
Dickinson  High  School  and  serves  to  illustrate  the  construc- 
tion and  operation  of  the  prony  brake.  Referring  to  Fig. 
84,  M  is  an  electric  motor  rated  to  give  2  hp.  at  1,500  r.p.m. 
and  108  volts.  In  order  to  check  up  the  rating,  the  motor  is 


94 


MECHANICS 


fitted  with  a  hollow  iron  pulley  (P),  which  is  rotated  against 
the  friction  of  the  adjustable  brake  (B).  The  brake  is  lined 
with  some  heat  resisting  material,  such  as  is  used  for  brake 
lining  of  an  automobile.  It  carries  an  arm  (A)  which  rests 
on  a  knife  edge  (K).  The  knife  edge  rests  on  the  platform 
balance  (J).  The  power  is  turned  on  and  the  thumb  screws 
(S  and  $')  are  tightened  until  the  speed  of  the  motor  is  approxi- 
mately 1,500  r.p.m.  in  a  clockwise  direction.  Meanwhile,  a 
stream  of  water  is  directed  against  the  pulley  to  keep  it  from 
overheating.  From  the  following  final  readings,  the  horse- 
power was  computed. 

Force  (F)  exerted  by  arm  on  balance 7  Ib. 

Length  (L)  of  arm 1  ft. 

Revolutions  per  minute  (r.p.m.) 1,530 

F  X  2L  X  3. 1416  X  r.p.m. 
B'hp'  =  33,000 

D,  7  X  2X3. 1416  X  1,530 

B'kp'=  33,000  =2.04hp. 

It  will  be  noted  that  the  numerator  of  the  above  fraction 
is  the  foot-pounds  of  work  which  it  is 
assumed  the  force  (F)  would  do  each 
minute,  in  rotating  through  a  distance 
of  2L  X  3.1416  every  revolution  for  1,530 
revolutions. 

Another  form  of  the   prony  brake  is 
shown    in    Fig.    85.     B    represents    the 
balance    wheel   of   an   automobile   from 
which  the  clutch  has  been  removed.     A 
belt  of  heat  resisting  material  is  carried 
around  the  wheel  and  secured  at  either 
FIG.  85.— Belt  form  of  end  by  two  spring  balances  (Ti  and  T2). 
prony  brake.          The  throttle  is  opened  and  the  balances 

tightened  until  the  wheel  is  making  1,000  r.p.m.  The  balance 
readings  are  taken  and  the  difference  in  tension  noted.  The 
wheel  is  kept  from  overheating  as  in  the  previous  experiment. 
From  the  following  data  the  horsepower  is  computed. 


POWER  95 

Diameter  (d)  of  wheel 1.5  ft. 

Difference  in  tension  (T2  -  T,) 200  Ib. 

Revolutions  per  min.  (r.p.m.) 1,000 

Thickness  of  belt 25  in. 

(T2  -  Ti)  X  d  X  3.1416  X  r.p.m., 

33,000 

200  X  1.521  X  3.1416  X  1,000      OA  , 
B'kp-  =  33,000  =  2°  hp' 

111.  S.  A.  E.  Determination  of  Horsepower. — The  deter- 
mination of  horse  power  by  the  S.  A.  E.  (Society  of  Auto- 
motive Engineers)  formula  is  purely  an  arbitrary  calculation 
and  is  used  only  in  computing  the  horsepower  of  automobile 
motors.  The  S.  A.  E.  method  is  derived  from  the  brake 
method  and  has  been  useful  for  comparative  purposes  such  as 
the  amount  paid  for  license  fee,  etc. 

D     =  bore  of  cylinders  in  inches, 
N     =  number  of  cylinders, 
2.5  =  a  constant. 


.       ,  .  , 
HP-  =  -S-K~  m  which, 


The  above  formula  assumes  a  piston  speed  of  1,000  ft.  per 
min.,  a  mean  effective  cylinder  pressure  of  90  Ib.  per  sq.  in., 
and  a  mechanical  efficiency  of  75  per  cent.  Since  a  gasoline 
motor  delivers  more  power  at  a  higher  rate  of  speed  and  since 
the  average  automobile  motor  has  a  piston  speed  of  1,500 
ft.  per  min.,  it  is  evident  that  the  S.  A.  E.  formula  is  more  or 
less  of  a  makeshift.  To-day  many  motors  are  tested  by  the 
brake  method  before  leaving  the  factory. 

112.  Indicated  Horsepower  (i.hp.). — The  indicated  horse- 
power method  is  confined  almost  exclusively  to  the  steam 
engine,  although  it  may  be  used  for  internal  combustion 
engines.  It  determines  the  power  developed  in  the  cylinder 
and  not  the  power  transmitted.  Indicated  horsepower  will 
always  exceed  brake  horsepower,  as  it  makes  no  deduction  for 
frictional  losses. 

In  finding  i.hp.,  it  is  necessary  to  make  use  of  work  dia- 
grams as  illustrated  in  Fig.  83.  The  diagrams  are  taken  by  a 

1  The  thickness  of  the  belt  is  added  to  the  diameter. 


96  MECHANICS 

special  device  called  an  indicator  and  from  the  diagrams  the 
mean  effective  pressure  is  found  for  the  stroke.  For  the 
actual  method  of  taking  indicator  diagrams,  the  student  is 
referred  to  books  dealing  with  steam  and  gas  engines. 

P  =  mean  effective  pressure  in  Ib. 


, ,          PLAN  .       ,.  , 
LhP-=3WWmwhlch> 


per  sq.  in., 

L  =  length  of  stroke  in  ft., 
A  =  Area  of  piston  head  in  sq.  in. 
N  =  Working  strokes  per  min. 

In  double  acting  engines  the  head  and  crank  ends  must  be 
figured  separately  and  added  together  for  the  total  horse- 
power. 

113.  Mechanical  Efficiency. — The  mechanical  efficiency  of 
an  engine  is  the  ratio  of  its  brake  horsepower  and  indicated 
horsepower. 

,,  brake  horsepower 

M.e.  = 


indicated  horsepower 

Questions  and  Problems 

1.  Define  and  give  an  example  of  power. 

2.  Give  the  general  formula  for  power. 

3.  Define  (a)  horsepower;  (6)  watt;  (c)  kilowatt. 

4.  State  the  mathematical  relation  between  the  above  units. 
6.  State  three  different  methods  of  determining  horsepower. 

6.  Describe  in  detail  two  different  ways  of  determining  b.hp. 

7.  What  is  the  8.  A.  E.  formula?     Is  it  accurate?     Why? 

8.  State  and  explain  the  formula  for  i.hp. 

9.  Why  is  b.hp.  less  than  i.hp? 

10.  How  may  the  mechanical  efficiency  of  an  engine  be  determined? 

11.  A  man  weighing  175  Ib.  climbs  a  ladder  through  a  vertical  distance 
of  30  ft.  in  1  min.     What  hp.  did  he  expend? 

12.  A  Goulds  triplex  pump  lifts  40  Ib.  of  water  a  vertical  distance  of 
5  ft.  in  1  min.     Compute  power  expended. 

13.  A  hoisting  engine  (gears  80  per  cent,  efficient)  is  used  in  lifting 
cement  to  the  top  of  a  building  50  ft.  high.     If  1,000  tons  is  to  be  lifted 
through  a  working  day  of  8  hr.,  what  is  the  average  horsepower  expended? 

14.  The  difference  in  tension  between  two  sides  of  a  belt  is  50  Ib.     If 
the  belt  is  running  at  the  rate  of  2,500  ft.  per  minute,  what  horsepower 
is  being  transmitted?     Express  the  answer  also  in  watts  and  kilowatts. 

15.  The  Cole  "Aero  Eight"  develops  horsepower  in  excess  of  80. 


POWER  97 

If  the  cylinders  are  33^  in.  bore,  what  is  the  horsepower  according  to  the 
S.  A.  E.  rating? 

16.  An  automobile  motor  has  an  indicated  hp.  of  35  and  a  brake  hp.  of 
30.     What  is  the  mechanical  efficiency  of  the  motor? 

17.  In  computing  the  power  of  an  electric  motor  by  a  prony  brake, 
the  following  data  was  obtained:  net  force  of  brake  arm  on  platform 
balance  11  lb.;  length  of  brake  arm  14  in.;  r.p.m.  1100.     Find  horsepower 
and  kilowatts  transmitted  by  the  motor. 

18.  A  belt  prony  brake  is  used  in  figuring  the  hp.  of  a  gasoline  motor. 
From  the  following  data  compute  the  power  transmitted  by  the  motor: 
difference  in  tension  of  balances  =  100  lb.;  diameter  of  pulley  =  2  ft.; 
r.p.m.  =  1,500. 

19.  In  determining  the  horsepower  of  the  Corliss  steam  engine  in  the 
Dickinson  High  School  power  plant,  a  group  of  students  reported  the 
following  data:  mean  effective  pressure  =  30  lb.  per  sq.  in.;  length  of 
stroke  =  24  in.;  diameter  of  piston  =  18  in.;  diameter  of  piston  rod  =  4 
in.;  r.p.m.  =  200.     Compute  the  indicated  horsepower  of  the  engine. 
(The  area  of  the  piston  rod  must  be  subtracted  from  the  area  of  the 
piston  on  the  "crank  end.") 


CHAPTER  XVI 
ENERGY 

114.  Energy. — Energy  is  the  ability  to  do  work.    A  body 
possessing  energy  may  or  may  not  do  work.     For  instance,  a 
stick  of  dynamite  possesses  a  large  amount  of  energy,  but  the 
energy  will  remain  latent  unless  the  dynamite  is  exploded. 
A  pound  of  coal  contains  over  10,000,000  ft.-lb.  of  energy.     No 
work  is  done,  however,  unless  the  coal  is  burned  and  the 
heat  energy  transformed  into  mechanical  energy,   as  in  the 
steam  power  plant.     Energy  should  never  be  confused  with 
force  or  power;  they  are  entirely  different  terms. 

115.  Fixed  Energy. — Energy  stored  up  in  a  body  is  called 
^  fixed   energy.     Food    contains    stored 

energy  from  which  the  body  derives 
its  energy  and  ability  to  do  work. 
Gasoline  contains  stored  energy,  every 
pound  furnishing  about  15,000,000  ft.-lb. 
Fixed  energy  of  this  kind  is  determined 
by  burning  the  substance  and  measuring 
the  heat  energy  given  off  during  the 
combustion. 

Ground  Many  bodies  possess  energy  on  account 
""""/  of  their  position.  This  form  of  fixed 
energy  is  called  potential  energy.  The 
pile  driver  furnishes  a  good  illustration. 
The  heavy  weight,  or  driver  when  released,  will  fall  toward 
the  earth  and,  on  striking  an  object,  will  accomplish  work. 
A  body  weighing  100  Ib.  and  suspended  10  ft.  above  the 
earth  has  1,000  ft.-lb.  of  potential  energy  (see  Fig.  86). 
In  order  to  elevate  the  body  it  was  necessary  to  do  1,000 
ft,-lb.  of  work.  The  potential  energy  possessed  by  any 

98 


ENERGY  99 

suspended  body  is  measured  by  the  amount  of  work  necessary 
to  raise  the  body  to  its  position. 

Potential  energy  =  Weight  X  height  =  W  X  h 
Potential  energy  may  also  be  due  to  elasticity.  If  a  coiled 
spring,  when  released,  is  able  to  exert  an  average  force  of 
100  Ib.  through  a  distance  of  4  in.,  it  is  evident  that  400  in.-lb. 
of  work  have  been  done.  It  is  likewise  evident  that  the  spring 
must  have  possessed  400  in.-lb.  of  potential  energy.  The 
valve  springs  of  a  gasoline  motor  furnish  a  good  example  of 
potential  energy  due  to  elasticity. 

Potential  energy  =  Force  X  distance  —  F  X  d. 

116.  Kinetic  Energy. — Kinetic  energy  is  the  energy  of  motion. 
A  rotating  fly  wheel  or  a  moving  projectile  possesses  kinetic 
energy.  In  fact,  any  moving  body  possesses  kinetic  energy, 
because  it  has  the  ability  to  do  work.  The  same  amount  of 
work  will  be  necessary  to  bring  a  body  to  rest  as  was  neces- 
sary to  set  the  body  in  motion.  If  a  motor  boat  has  50,000,- 
000  ft.-lb.  of  kinetic  energy,  50,000,000  ft.-lb.  of  work  will 
be  done  in  bringing  the  boat  to  a  stop. 


„.    ..  WV* 

Kinetic  energy  =  — ~ — 

ft.  Ib.,  in  which 
Kinetic  energy    =    1/2 
M  V2  ergs,  in  which 


W  =  weight   of   the   body  in  Ib., 
V  =  velocity  per  sec.  in  ft., 
g  =  32.16,  acceleration  of  gravity. 

M  =  mass  of  the  body  in  grams, 
V  =  velocity  per  sec.  in  cm. 


117.  Why  a  Body  Possesses  Energy. — Every  body  possessed 
of  energy  has,  at  some  time,  had  work  done  upon  it.  A  com- 
pressed spring  has  potential  energy,  due  to  the  fact  that  work 
was  done  in  compressing  the  spring.  A  steam  shovel,  sus- 
pended above  the  ground,  possesses  potential  energy,  due 
to  the  fact  that  a  certain  amount  of  work  was  done  upon  the 
shovel  in  lifting  it.  Similarly,  a  motor  cycle,  coasting  along 
a  level  stretch,  owes  its  kinetic  energy  to  the  fact  that  work 
was  done  in  imparting  to  it  the  momentum.  Work,  it  will 
be  seen,  consists  of  taking  energy  from  one  body  and  giving  it 
to  another  body. 


100  MECHANICS 

118.  Transformation  of  Energy. — If  the  weight  shown  in 
Fig.  86  is  released,  it  will  have  1,000  ft.-lb.  of  kinetic  energy 
on  striking  the  ground.     During  the  fall  the  potential  en- 
ergy   was    being    converted    into    kinetic    energy.     At  any 
time  during  the  downward  course  of  the  weight,  the  weight 
possessed  1,000  ft.-lb.  of  energy,  that  is,  the  sum  of  the  poten- 
tial  and  kinetic  energy  was  constant.     Starting  with  poten- 
tial energy  alone,  we  see  a  gradual  loss  of  potential  energy 
and  a  proportionate  increase  of  kinetic  energy  until,  on  strik- 
ing the  ground,  the  body  has  entirely  lost  its  original  potential 
energy  and  possesses  a  like  amount  of  kinetic  energy. 

Other  transformations  of  energy  may  be  noted  such  as: 
(1)  Transformation  of  heat  energy  into  mechanical  energy,  as 
in  the  steam  engine;  (2)  transformation  of  mechanical  energy 
into  electrical  energy,  as  in  the  steam-electric  plant;  and 
transformation  of  mechanical  energy  into  heat  energy,  illus- 
trated in  stopping  an  automobile  by  means  of  brakes.  Other 
examples  might  be  given,  but  these  are  sufficient  to  show  that 
one  kind  of  energy  may  be  replaced  by  another. 

119.  Conservation  of  Energy. — According  to  the  law  of 
conservation  of  energy,  the  amount  of  energy  in  the  universe  is 
constant.     In   other   words,   energy  may  neither  be   created 
nor  destroyed.     We  have  seen,  however,  that  energy  may  be 
transformed  from  one  kind  to  another.     Useful  energy,  which 
disappears  as  such,   reappears  in  some  other  form  and  is 
called  dissipated  energy.     The  electrical  energy  of  an  incandes- 
cent light  bulb  is  transformed  into  light  and  heat  energy.     Since 
the  heat  energy  of  no  use,  it  is  called  dissipated  energy.     Yet 
in  no  way  has  the  original  electrical  energy  been  destroyed. 
Likewise  the  heat  and  light  energy  was  not  created ;  it  resulted 
from  the  transformation  of  the  electrical  energy. 

120.  The   Mechanical  Equivalent  of  Heat.— In  order  to 
measure  the  amount  of  heat  possessed  by  a  body,  a  unit 
known  as  the  British  thermal  unit  is  employed.     The  British 
thermal  unit  (B.t.u.)  is  the  amount  of  heat  necessary  to  raise 
1  Ib.  of  water  1  degree  Fahrenheit  in  temperature.     If  10  Ib.  of 
water  are  raised  4°  Fahrenheit  in  temperature,  then  40  B.t.u. 


ENERGY  101 

of  heat  energy  was  necessary  to  effect  the  rise.  It  has  been 
found  by  experiment  that  there  is  a  strict  relation  between 
work  done  and  heat  generated.  It  has  been  determined  that 
780  ft.-lb.  of  work  always  result  in  the  appearance  of  1  B.t.u. 
of  heat;  or,  conversely,  that  1  B.t.u.  of  heat  will  produce  780 
ft.-lb.  of  work.  This  relation  was  discovered  by  James  Pres- 
cott  Joule,1  in  whose  honor  the  joule  was  named. 

1  James  Prescott  Joule  (1818-1889).  Famous  English  physicist  and 
chemist.  Determined  the  mechanical  equivalent  of  heat  and  estab- 
lished the  law  of  conservation  of  energy.  Made  many  valuable  experi- 
ments in  electricity  and  magnetism. 

Questions  and  Problems 

1.  Define  energy.     Explain  why  a  body  possesses  energy. 

2.  What  is  fixed  energy?     Give  examples. 

3.  What  is  potential  energy?     Give  examples  of  potential  energy 
due  to  (a)  position;  (6)  elasticity. 

4.  How  is  the  energy  in  a  pound  of  coal  measured? 
6.  Give  the  formulas  for  potential  energy. 

6.  What  is  kinetic  energy?     Give  examples. 

7.  Give  the  formulas  for  kinetic  energy. 

8.  Give  several  illustrations  of  the  transformation  of  energy. 

9.  State  and  discuss  the  law  of  conservation  of  energy. 

10.  What  is  meant  by  the  mechanical  equivalent  of  heat.     Discuss 
fully. 

11.  A  mine  cage  weighing  500  Kg.  is  suspended  at  the  top  of  a  shaft 
100  m.  deep.     What  is  its  potential  energy  with  respect  to  the  bottom  of 
the  shaft? 

12.  How  much  kinetic  energy  will  the  cage  in  the  previous  problem 
have  on  striking  if  the  cable  breaks? 

13.  A  coiled  spring,  when  released,  exerts  a  pressure  of  50  Ib.  through 
a  distance  of  3  in.     What  potential  energy  did  it  possess? 

14.  A  ton  pile  driver  drops  10  ft.     What  is  the  average  resistance  of 
the  pile  if  it  is  driven  down  2  in? 

16.  A  truck  weighing  2  tons  is  moving  at  the  rate  of  20  m.p.h.  What 
force  must  be  applied  to  stop  it  in  300  ft.  if  the  engine  is  shut  off? 

16.  What  is  the  kinetic  energy  of  a  tug  weighing  100  tons  and  moving 
at  the  rate  of  10  m.p.h J     How  much  work  must  be  done  to  bring  the 
tug  to  a  stop  if  the  steam  is  shut  off? 

17.  A  projectile  weighing  500  Kg.  has  a  muzzle  velocity  of  1,000m.  per 
sec.     What  is  its  kinetic  energy? 

18.  A  baseball  weighing  5  oz.  strikes  the  catcher's  mitt  with  a  velocity 
of  40  ft.  per  sec.     What  is  the  kinetic  energy  on  striking? 


102  MECHANICS 

19.  An  automobile  weighing  3,000  Ib.  is  driven  at  the  rate  of  50  m.p.h. 
on  a  level  stretch.     The  power  is  shut  off  just  as  the  car  strikes  a  3  per 
cent,  grade.     How  far  up  grade  will  it  go  before  stopping?     Assume  no 
loss  by  friction. 

20.  A  spike  is  being  driven  by  a  sledge.     The  sledge  weighs  40  Ib. 
and  strikes  the  spike  at  the  speed  of  10  ft.  per  sec.     If  the  spike  is  driven 
in  1  in.  at  each  blow,  what  average  resistance  does  it  work  against? 


CHAPTER  XVII 
FRICTION 

121.  Friction. — Friction  is  the  resistance  offered  to  the  move- 
ment of  one  body  over  another.     This  resistance  is  due  to  the 
fact  that  no  surface  is  perfectly  smooth.     The  microscope 
reveals  to  us  that  every  body,  no  matter  how  smooth  it  may 

seem,  is  covered  with  countless    

small  projections.     When  bodies 

are  in  contact,  these  projections 
interlock,  giving  rise  to  friction. 
Suppose  we  take  as  an  illustra-  T 

FIG.    87. — Diagram    (exaggerated) 
tion    two    Surfaces    which    have  showing  cause  of  friction. 

been  planed  and  polished  with 

the  utmost  care.  When  examined  under  the  microscope,  the 
surfaces  are  found  to  be  made  up  of  "hills  and  valleys"  as 
shown  (with  exaggeration)  in  Fig.  87.  It  will  be  seen  that  the 
projections  of  A  sink  into  the  depressions  of  B.  In  order  to 
move  A  horizontally  along  B,  it  is  necessary  to  lift  the  upper 
surface  of  A  until  the  projections  cease  to  interlock.  Lifting 
a  body  involves  work.  Since  work  involves  force,  it  is  easy  to 
see  why  force  is  necessary  in  overcoming  friction. 

122.  Advantages  and  Disadvantages  of  Friction. — It  would 
be  impossible  for  us  to  do  without  friction.     If  there  was  no 
friction,  we  would  be  unable  to  walk  or  stand;  it  would  be 
almost  impossible  for  us  to  pick  up  objects.    Locomotives 
would  be  useless,  as  there  would  be  no  traction  between  the 
drivers  and  rails.     Belt  driven  machines  and  friction  clutches 
would  not  operate. 

On  the  other  hand,  friction  is  a  perplexing  problem  for  the 
engineer.  He  is  constantly  endeavoring  to  eliminate  it, 
because  it  impairs  the  efficiency  of  machines,  resulting  in  wear 

103 


104 


MECHANICS 


and  dissipated  power.  Power  transmitted  through  spur  or 
bevel  gears  is  cut  down  markedly  by  friction.  A  worm-driven 
automobile,  noticeable  for  silent  running,  is  impaired  in 
efficiency  by  the  friction  of  the  worm  and  worm  gear.  It  is 
evident,  from  the  above  discussion,  that  friction  is  desirable 
in  certain  ways  and  undesirable  in  others. 

123.  Coefficient  of  Friction. — The  fraction  representing  the 
relation  between  friction  and  pressure  is  known  as  the  coeffi- 
cient of  friction. 

static  friction 


Coefficient  of  static  friction     = 


Coefficient  of  kinetic  friction  = 


pressure  normal  to  surfaces 

kinetic  friction 
pressure  normal  to  surfaces 


FIG.  88. — Apparatus  for  determining  coefficient  of  sliding  friction. 


The  coefficient  of  sliding  friction  only  will  be  considered  here 
and  is  usually  spoken  of  simply  as  the  coefficient  of  friction. 
It  may  be  determined  by  means  of  the  apparatus  shown  in 
Fig.  88.  ABC  is  an  adjustable  inclined  plane  to  which  may 
be  attached  various  substances  such  as  sheet  copper,  leather, 
brake  linings,  etc.  Suppose  we  wish  to  determine  the  coeffi- 
cient for  carbon  and  copper.  The  copper  sheet  is  firmly 
attached  to  the  upper  surface  of  the  plane  and  a  4  Ib.  carbon 
block  is  placed  on  the  copper.  The  plane  is  now  adjusted 
until  the  block  just  refrains  from  moving  down  the  plane,  The 


FRICTION  105 

block  is  gently  tapped  and  the  angle  0  slowly  increased  by  the 
fine  adjustment  J  until  the  carbon  begins  to  move  slowly 
and  uniformly  down  the  copper  sheet.  The  coefficient  of 
friction  is  equal  to  the  height  of  the  plane  divided  by  the  base 
of  the  plane. 

Coefficient  of  friction  =  -j~ 

The  height  and  length  are  measured  and  are  found  to  be 
6  in.  and  24  in.  respectively.  Hence  the  coefficient  for  carbon 
and  copper  will  be  6/24  =  .25.  Since  height/base  equals  the 
tangent  of  angle  6,  it  is  evident  that  the  coefficient  is  also 
equal  to  the  tangent  of  angle  6  (in  this  case  14°) . 

Coefficient  of  friction  =  tangent  of  angle  0 

By  referring  to  Fig.  88,  the  above  mathematical  relations 
are  easily  explained.  W,  the  weight  of  the  block,  may  be 
resolved  into  two  components :  Fr.  or  the  force  parallel  to  the 
plane  just  necessary  to  overcome  friction  and  Pr.  the  per- 
pendicular pressure  between  the  surfaces.  The  coefficient 
is  evidently  Fr./Pr.  By  similar  triangles  it  may  be  proven  that 
Fr.  is  proportional  to  AC  or  the  height  of  the  plane,  and  that 
Pr.  is  proportional  to  BC  or  the  base  of  the  plane.  Hence  the 
coefficient  is  equal  to  the  height/base  or  the  tangent  of  angle  6. 

124.  Laws  of  Friction. — There  are  so  many  elements  enter- 
ing into  a  consideration  of  friction  that  it  almost  impossible  to 
formulate  any  exact  laws  concerning  it.  Friction  depends  on 
materials,  pressure,  smoothness,  amount  and  kind  of  lubrica- 
tion, temperature,  etc.  The  following  laws  are  accurate 
enough  for  ordinary  purposes. 

FOB  SOLIDS: 

1.  Friction  varies  with  the  nature  of  the  surfaces  in  contact. 

2.  Friction  is  proportional  to  the  perpendicular  pressure  between  the 
surfaces  in  contact. 

3.  Friction  is  not  affected  by  the  amount  of  the  area  of  the  surfaces  in 
contact. 

4.  Friction  decreases  somewhat  as  the  velocity  increases. 


106 


MECHANICS 


FOR  FLUIDS: 

1.  Except  at  high  speeds,  friction  is  little  affected  by  the  nature  of  the 
surfaces  in  contact. 

2.  Friction  is  independent  of  the  pressure  between  the  surfaces  in  contact. 

3.  Friction  depends  upon  the  area  of  contact  and  is  proportional  to  it. 

4.  Friction  increases  with  velocity  and  is  very  great  at  high  speeds. 

123.  Anti-friction  Devices. — The  modern  machine  has  been 
greatly  increased  in  efficiency  through  the  employment  of 
anti-friction  devices.  The  simplest  method  of  reducing  friction 
is  to  make  the  surfaces  in  contact  as  smooth  as  possible  by 
mechanical  means,  such  as  planing  and  polishing.  The  fric- 
tion is  further  reduced  if  the  surfaces  are  lubricated.  Bear- 
ings  are  used  to  cut  down  the  friction  of  rotating  bodies.  A 
bearing  is  a  device  by  which  rotating  bodies  are  held  in  place. 
They  are  classed  as  low,  medium  or  high  duty,  according  to  the 
load  they  are  able  to  carry.  Bearings  are  either  plain,  in 
which  one  surface  rubs  against  another;  ball;  or  roller.  If  a 
bearing  is  designed  to  take  care  of  a  load  at  right  angles  to 
the  rotating  body,  it  is  called  a  radial  bearing;  if  it  takes  care 
of  a  load  longitudinal  to  the  rotating  body,  it  is  called  a  thrust 
bearing.  Many  bearings  are  a  combination  of  the  two. 

126.  Plain  Bearings. — The  accompanying  diagram  (Fig. 
89)  shows  the  connecting  rod  bearings  of  a  "  Universal  Una- 


FIG.  89. — Steam  engine  connecting  rod  with  piain  bearings. 

flow"  steam  engine.  The  rod  is  made  from  forged  steel  and 
fitted  with  a  heavy  iron  box  lined  with  babbitt  at  the  crank 
end  and  a  bronze  box  at  the  crosshead  end.  Wear  is  taken 
care  of  by  means  of  the  adjustments  shown.  Babbitt  is 
a  light,  whitish  material  and  is  an  alloy  of  tin,  copper  and 
antimony.  It  has  a  low  melting  point  and  is  very  easy  to 


FRICTION 


107 


work.     Plain  bearings  are  used   almost  exclusively  on  the 
steam  engine. 

Figure  90  shows  the  bearings  used  on  the  connecting  rod  of  a 
Cole  " Aero-Eight"  automobile  and  is  characteristic  of  auto- 
mobile connecting  rods  in  general.  The  rod  is  joined  to  the 


FIG.  90. — Diagram  showing  plain  bearings  used  on  the  connecting  rod  of  an 

automobile. 

crankshaft  by  a  babbitt  lined  bearing.  It  is  secured  to  the 
piston  by  a  steel  wrist  pin  forced  into  place  and  employs  a 
bronze  bushing  as  a  bearing.  Bronze  outwears  babbitt,  but 
is  not  so  easily  worked. 

127.  Ball  Bearings. — Ball  bearings  are  used  on  magnetos, 
electric    motors    and    generators,    automobile    transmissions, 


FIG.  91. — Cup-and-cone  ball 
bearing. 


FIG.  92. — Annular  ball  bearing. 


airplane  propellers,  etc.  The  cup-and-cone  bearing  (Fig.  91) 
is  designed  to  stand  occasional  end  thrusts  and  is  adjustable. 
The  cup  and  cone,  however,  must  be  set  at  an  angle  which 


108 


MECHANICS 


lessens  the  radial  load  capacity.     The  efficiency  of  this  type 
of  bearing  is  greatly  reduced  when  wear  occurs. 

The  annular  ball  bearing  (Fig.  92)  has  a  heavy  vertical  load 
capacity,  but  makes  no  provision  for  thrust.  If  thrust  loads 
are  present  to  any  extent,  thrust  bearings  must  be  used  in 


FIG.  93. — Fafnir  radial  ball 
bearing  assembled. 


FIG.  94. — Fafnir  radial  ball  bearing 
disassembled. 


FIG.  95. — Fafnir  thrust  ball 
bearing  (flat-race  type)  unas- 
sembled. 


FIG.  96. — Fafnir  thrust  ball  bearing  (self- 
aligning  type)  unassembled. 


addition.  The  annular  type  of  bearing  is  not  adjustable  and 
must  be  replaced  after  wear.  It  is  more  correct  in  design 
than  the  cup-and-cone  bearing  and  is  preferred  for  that 
reason. 

Fafnir  ball  bearings  are  representative  of  the  best  and  are 
here  selected  for  description.  The  student  should  examine 
the  illustrations  above  carefully. 


FRICTION 


109 


Fafnir  bearings  are  made  of  an  alloy  of  high  carbon,  chrome 
steel,  hardened  by  a  special  heat  treatment  and  quenched  in 
oil  to  insure  proper  density  and  hardness.  The  raceways  are 
very  accurately  ground  and  the  balls  are  guaranteed  round  and 
true  to  1/10,000  of  an  inch. 

128.  Roller  Bearings. — Roller  bearings  have  proven  very 


Inner  race  Roller  assembly  Outer  race 

FIG.  97. — Hyatt  roller  bearing  unassembled. 


FIG.  98. — Timken  roller  bearing 
partly  unassembled. 


FIG.  99. — Front  wheel  of  an 
automobile  mounted  on  Timken 
bearings. 


satisfactory,  particularly  for  medium  and  heavy  loads.  The 
Hyatt  and  Timken  bearings  are  selected  for  study  on  account 
of  their  wide  use. 

The  Hyatt  bearing  (Fig.  97)  is  used  for  automobiles,  mine 


110 


MECHANICS 


FIG.    100. — Timken  bearings  used   on  a 
worm-drive  truck. 


cars,  line  shafts,  refrigerating  machines,  printing  presses,  etc. 
It  consists  of  flexible,  hollow  rollers  of  heat-treated  alloy  steel 
wound  helically  and  an  inner  and  outer  race.  If  shafts  are 

sufficiently  hard,  the  inner 
race  may  be  dispensed 
with.  The  inner  race  is 
made  from  seamless  tubing 
of  alloy  steel,  carburized 
and  heat  treated.  The 
outer  race  is  of  the  same 
material  and  is  either  split 
or  solid.  For  heavy  work, 
the  solid  type  should  al- 
ways be  used.  The  rollers 
are  held  in  place  by  heavy 
bars  riveted  to  the  end 
rings.  The  bars  and  rings  constitute  the  cage.  Hyatt  bear- 
ings make  no  provision  for  thrust. 

The  Timken  bearing  (Fig.  98)  has  a  ribbed  cone  of  carbon- 
ized, electric  steel,  with  outside  taper;  tapered  rollers  of  electric 
alloy  steel,  heat-treated;  a  tapered  cage  or  roller  retainer  of 
sheet  steel ;  and  a  cup  or  outer  race  of  electric  steel,  with  inside 
taper.  It  will  be  observed  that  this  bearing  is  a  combination 
thrust  and  radial  bearing;  also  that  its  peculiar  construction 
enables  wear  to  be  taken  up  easily. 

Timken  bearings  are  used  extensively  for  automobiles  at 
points  of  hard  service  such  as  rear  and  front  wheels,  differential, 
pinion  or  worm  and  transmission.  Figures  99  and  100  show 
installations  of  Timken  bearings  on  the  automobile. 

129.  Lubrication  of  Bearings. — The  best  lubricant  for 
bearings  is  a  neutral  mineral  oil  or  grease,  entirely  free  of  any 
acid,  alkali  or  sulphur.  Animal  and  vegetable  oils  are  apt  to 
contain  acid  in  sufficient  quantities  to  cause  corrosion.  A 
good  lubricant  will  not  only  lessen  friction,  but  will  also 
protect  the  metallic  surfaces  from  the  action  of  the  atmosphere 
and  aid  in  the  exclusion  of  dirt  and  water.  A  light  or  medium 
oil  is  generally  used  for  high  speeds,  while  liquid  grease  or 
heavy  machine  oil  is  used  for  slow  speeds. 


FRICTION  111 

Questions  and  Problems 

1.  Define  friction. 

2.  Show,  using  diagram,  why  force  is  necessary  in  overcoming  friction. 

3.  Discuss  the  advantages  and  disadvantages  of  friction. 

4.  What  is  static  friction?     Kinetic  friction? 

5.  What  is  meant  by  the  coefficient  of  friction?     Describe  a  method 
of  determining  it. 

6.  State  four  laws  for  friction  between  solids;  four  laws  for  fluid 
friction. 

7.  State  the  various  methods  employed  in  reducing  friction. 

8.  What  is  a  bearing?     Radial  bearing?     Thrust  bearing? 

9.  Describe  typical  plain  bearings  and  give  concrete  illustrations 
showing  where  they  are  used.     Discuss  the  relative  advantages  of  babbitt 
and  bronze. 

10.  Describe  the  ball  bearing.     For  what  kind  of  work  is  it  used? 

11.  What  is  an  annular  bearing?     Cup-and-cone  bearing? 

12.  Describe  the  Hyatt  roller  bearing.     What  is  the  advantage  of  the 
spiral  rollers? 

13.  Describe  the  Timken  roller  bearing.     What  is   the  distinctive 
feature  of  the  Timken  bearing? 

14.  A  force  of  6  Ib.  is  necessary  to  draw  a  sled  weighing  50  Ib.  along  a 
horizontal  surface.     Find  coefficient  of  friction  and  work  done  if  the  sled 
moves  50  ft. 

16.  The  coefficient  of  friction  for  oak  on  leather  is  .30.  What  force 
is  necessary  to  pull  an  oak  block  weighing  7  Ib.  slowly  and  uniformly  over 
a  level,  leather  surface? 

16.  If  it  takes  a  force  of  10  Ib.  to  slide  a  piece  of  ice  across  the  floor  and 
the  coefficient  of  friction  is  .06,  what  is  the  weight  of  the  ice? 

17.  An  inclined  plane  has  an  upper  surface  of  dry  agate.     The  plane  is 
adjusted  until  a  steel  block  weighing  10  Ib.  slides  slowly  and  uniformly 
down  the  plane.     Find  the  coefficient  of  friction  for  steel  on  dry  agate, 
if  the  plane  is  5  ft.  long  and  1  ft.  high. 

18.  The  experiment  was  repeated  with  oiled  agate  and  the  coefficient 
of  friction  was  found  to  be  .107.     What  was  the  height  of  the  plane? 

19.  An  iron  block  is  hauled  along  a  level  stone  surface  by  a  force  of 
200  Ib.  acting  at  an  angle  of  5°  to  the  horizontal.     If  the  coefficient  of 
fricton  is  .50,  what  is  the  weight  of  the  block? 


CHAPTER  XVIII 
SIMPLE  MACHINES 

130.  The  Machine. — The  machine  is  a  mechanical  device 
designed  to    do  work   advantageously.     A  machine   does  not 
possess  energy  in  itself,  but  must  receive  energy  from  some 
outside   source.     The   original   energy,    after  various  trans- 
formations and  dissipations,  will  be  delivered  in  part  to  some 
special  place  where  the  work  is  to  be  done.     It  should  be 
borne  in  mind  that  no  energy  is  ever  destroyed,  but  simply 
disappears  as  useful  energy.     The  great  problem  in  connec- 
tion with  machines  is  to  reduce  all  operating  losses  as  far  as 
possible;  that  is,  to  work  for  a  higher  efficiency  by  causing  the 
output  to  approach  the  input  more  closely. 

131.  Simple  Machines. — Simple  machines  are  divided  into 
six  general  classes,  according  to  the  principle  upon  which  they 
work;  namely,  the  lever,  pulley,  wheel  and  axle,  inclined  plane, 
screw  and  wedge.     As  will  be  seen  later,  there  are  in  reality 
only  two  different  classes,  namely,  the  lever  and  inclined  plane. 
The  pulley  and  wheel  and  axle  are  modified  levers,  and  the 
screw  and  wedge  are  modified  inclined  planes.     All  machines 
operate  on  the  principle  of  the  lever  or  inclined  plane  or  a 
combination  of  the  two.     It  will  be  assumed,  in  studying 
simple  machines,  that  no  friction  is  present  and  that  the  effi- 
ciency is  therefore  100  per  cent. 

132.  Important  Definitions. — In   discussing  machines,   we 
shall   have   occasion   to   refer   repeatedly   to   input,    output, 
efficiency,  velocity  ratio  and  mechanical  advantage.     A  proper 
study  of  machines  depends  upon  a  thorough  understanding 
of  the  above  terms. 

Input. — Input  is  the  energy  received  by  a  machine  from  some 
outside  source. 

112 


SIMPLE  MACHINES  113 

Output. — Output  is  the  energy  which  a  machine  is  capable 
of  delivering  and  is  always  less  than  the  input. 

Efficiency. — Efficiency  is  the  ratio  of  the  output  to  the  input. 
It  is  customary  to  reduce  the  decimal  thus  obtained  to  per  cent. 

^^  .  output 

Efficiency  =  -^ 

input 

Velocity  Ratio. — Velocity  ratio  is  the  distance  through  which 
the  driving  force  acts  divided  by  the  distance  through  which  the 
resisting  force  acts  in  the  same  time. 

Mechanical  Advantage. — Mechanical  advantage  is  the  ratio 
of  the  resisting  force  or  load  to  the  driving  force  or  effort. 

i\/r  A   —  resistin9  force  _  R 
driving  force       E 

133.  Law  of  Frictionless  Machines. — Provided  there  was  no 
friction,  the  following  law  would  apply  to  all  machines, 

The  driving  force  (E)  X  the  distance  (di)  through  which  it 
acts  =  the  resisting  force  (R)  X  the  distance  (d%)  through  which 
it  acts  in  the  same  time. 

E  X  di  =  R  X  d2 

134.  General  Law  for  All  Machines. — Since  friction  can 
never  be  eliminated  entirely,  for  practical  purposes  the  above 
law  should  be  amended  to  read, 

The  driving  force  (E)  X  the  distance  (di)  through  which  it 
acts  X  the  efficiency  (Effic.)  =  the  resisting  force  (R)  X  the 
distance  (dz)  through  which  it  acts  in  the  same  time. 

E  X  di  X  Effic.  =  R  X  d2 

Since  input  is  equal  to  output  plus  the  energy  used  in  over- 
coming friction,  the  law  may  also  be  stated, 

The  driving  force  (E)  X  the  distance  (di)  through  which  it 
acts  =  the  resisting  force  (R)  X  the  distance  (d%)  through 
which  it  acts  in  the  same  time  +  friction  (Fr.)  X  the  distance 
(c?3)  through  which  it  acts  also  in  the  same  time. 

E  X  di  =  (R  X  di)  +  (Fr.  X  d») 
It  is  essential  that  the  student  should  thoroughly  under- 

8 


114  MECHANICS 

stand  the  laws  given  in  this  paragraph.  These  laws  apply  to 
all  machines  (frictionless  or  non-f rictionless) .  Any  subse- 
quent "laws"  relating  to  machines  will  be  found  to  be  simply 
special  applications  of  the  general  law. 

General  Questions  Relating  to  Machines 

1.  What  is  a  machine?     What  is  the  source  of  energy  in  a  machine? 

2.  What  are  the  six  simple  machines?     Into  what  two  ultimate  classes 
may  they  be  divided? 

3.  What  is  meant  by  input,  output,  efficiency,  velocity  ratio,  mechan- 
ical advantage? 

4.  What  is  the  law  of  "frictionless"  machines?     Is  perpetual  motion 
possible?     Why? 

6.  What  is  the  "general  law"  of  all  machines?     State  the  law  in  two 
different  ways.     Show  that  this  law  applies  to  "frictionless"  machines. 
6.  How  may  the  efficiency  of  a  machine  be  increased? 

135.  The  Lever.1 — The  lever  is  a  rigid  rod  or  bar  designed  to 
rotate  about  a  fixed  point  called  the  fulcrum.     Unless  otherwise 
noted,    levers    will    be    considered    weightless.    Levers    are 
divided  into  three  classes,  according  to  the  relative  positions 
of  the  effort,  resistance  and  fulcrum.     A  lever  will  be  in 
equilibrium  when, 

The  effort  (E)  X  the  perpendicular  distance  (di)  to  the  ful- 
crum =  the  resistance  (R)  X  the  perpendicular  distance  (dz)  to 
the  fulcrum. 

E  X  di  =  R  X  dz 

136.  The  First  Class  Lever. — The  first  class  lever  is  illus- 


FIG.  101. — First  class  lever. 

trated  by  a  crow  bar.     A  heavy  weight  may  be  raised  by  the 
application  of  a  relatively  small  force.     An  examination  of 
Fig.  101  shows  that  the  fulcrum  is  between  the  effort  and 
1  It  is  suggested  that  Chap.  X  be  reviewed  at  this  point. 


SIMPLE  MACHINES 


115 


resistance  and  that  speed  is  sacrificed  for  force.  In  case  of 
rotation,  the  output  will  be.  equal  to  the  input  and  the  v.r. 
and  m.a.  will  be  numerically  the  same. 

137.  The  Second  Class  Lever.  —  The  second  class  lever  is 
illustrated  by  a  wheelbarrow.     An  examination  of  Fig.  102 


E--IOO 


r 


r 

N - 


FIG.   102. — Second  class  lever. 


shows  that  the  resistance  is  between  the  fulcrum  and  effort 
and  that  speed  is  sacrificed  for  force.  In  case  of  rotation,  the 
output  will  be  equal  to  the  input  and  the  v.r.  and  m.a.  will  be 
numerically  the  same. 

138.  The  Third   Class  Lever. — The  third   class  lever  is 
illustrated  by  the  fishing  rod.     An  examination  of  Fig.  103 

E*/00# 


-> 


Exd,  = Fix el? 


FIG.  103. — Third  class  lever. 

shows  that  the  effort  is  between  the  fulcrum  and  the  resistance 
and  that  force  is  sacrificed  for  speed.     In  case  of  rotation,  the 


116 


MECHANICS 


output  will  be  equal  to  the  input  and  the  v.r.  and  m.a.  will  be 
numerically  the  same. 

Questions  and  Problems  Relating  to  the  Lever 

1.  What  is  a  lever?     Describe  the  different  classes  of  levers  and  give 
an  example  of  each. 

2.  State  the  law  of  "frictionless,"  "weightless"  levers.     Show  that 
this  law  is  an  application  of  the  general  law  of  machines. 

3.  In  case  the  lever  is  not  considered  weightless,  how  would  the  above 
law  read? 


450* 


FIG.  104. 


4.  State  the  class  of  lever  illustrated  in  the  following  cases:   (a) 
drawing  a  nail  with  a  claw  hammer;  (6)  rowing  a  boat;  (c)  turning  a 
grindstone  by  foot-power;  (d)  using  a  pair  of  shears;  (e)  lifting  a  pail  of 
water;  (/)  moving  the  spark  control  of  an  automobile;  (g}  cranking  an 
automobile. 

5.  Determine  the  velocity  ratio  and  mechanical  advantage  in  Figs. 
101,  102  and  103;  also  the  input  and  output.     Assume  a  rotation  of  360°. 

6.  A  first  class  lever  is  10  ft.  long.     Where  must  the  fulcrum  be  placed 
in  order  that  a  weight  of  100  Ib.  suspended  at  one  end  may  be  balanced 
by  a  weight  of  50  Ib.  at  the  other  end? 

7.  A  second  class  lever  is   10  ft.  long. 
Where  must  a  weight  of  100  Ib.  be  placed 
in  order  to  balance  a  force  of  50  Ib.  acting 
at  the  end  of  the  lever? 

8.  A  third  class   lever    is    10   ft.    long. 
What  force  applied  6  ft.  from  the  fulcrum 
will  balance  a  weight  of  100  Ib.  at  the  oppo- 
site end? 

9.  What  force  applied  at  E  (Fig.  104)  will  produce  equilibrium? 
10.  A  pressure  of  25  Ib.  is  applied  to  the  emergency  brake  lever  of  a 
Studebaker  automobile.     Jf  the  resistance  arm  is  2%  in.  and  the  effort 
is  applied  13>£  in.  from  the  fulcrum,  what  is  the  tension  in  the  brake  rod? 
What  is  the  m.a.?    v.r.f 


W 


FIG.  106. 


SIMPLE  MACHINES 


117 


11.  Find  the  .tension  in  rod  B  for  the  treadle  as  shown  in  Fig.  105. 
What  force  acts  on  the  pin  at  C?     What  is  the  m. a.  f    v.r? 

12.  Figure  106  represents  a  lever  safety  valve  used  on  stationary 
steam  engines.     The  valve  is  2  in.  in  diameter  and  weighs  3  Ib.     Assum- 
ing the  lever  to  be  weightless,  what  weight  must  be  suspended  at  W  so 
that  the  valve  will  "blow"  at  a  pressure  of  175  Ib.  per  sq.  in.? 

139.  Pulleys.1 — The  pulley  is  a  wheel  (generally  of  wood  or 
metal)  having  a  grooved  circumference  and  able  to  rotate  freely 
about  a  fixed  axis  at  its  center.  If  several  pulleys  are  combined 
along  the  same  axis  in  the  same  sheath  which  is  fixed,  and  if  a 


J00# 


FIG.   107.—  Single    fixed 
pulley. 


FIG.  108.—  Single  mov- 
able  pulley. 


too* 


FIG.  109.—  Single  fixed 
and  single  movable  pui- 
ley. 


like  combination  which  is  movable  is  joined  to  it  by  means  of  a 
rope  or  chain,  we  have  what  is  known  as  a  block  and  tackle. 
In  principle  the  pulley  is  a  rotating  lever.  It  may  bring 
about  a  gain  in  force  or  an  advantage  of  direction.  Pulleys 
are  usually  arranged  horizontally  in  the  sheath;  for  con- 
venience, the  vertical  arrangement  is  used  here. 

The  single  fixed  pulley  (Fig.  107)  has  an  advantage  of  direc- 
tion only,  E  and  R  being  equal  since  the  tensions  T\  and  Tz 
are  100  Ib.  each.  The  m.a.  is  100/100  or  1.  E  and  R  have 
the  same  speed  in  case  of  rotation;  hence  the  v.r.  is  1. 

The  single  movable  pulley  (Fig.  108)  is  used  when  a  gain  in 
force  is  desired.  The  supporting  tensions  TI  and  T2  are 
50  Ib.  each  and  the  m.a.  is  100/50  or  2.  Since  E  has  twice  the 
velocity  of  R,  the  v.r.  will  be  2. 

1  The  pulleys  discussed  here  are  considered  to  be  weightless. 


118 


MECHANICS 


A  single  fixed  pulley  is  generally  used  with  a  single  movable 
pulley  (Fig.  109).  The  extra  pulley  enables  the  effort  to  be 
applied  more  conveniently.  The  m.a.  and  v.r.  are  the  same 
as  in  the  previous  case. 

Figure  110  shows  another  arrangement  of  a  single  fixed 
and  single  movable  pulley.  The  supporting  tensions  TI  and 


ft 

IOO 

FIG.  110. — Single 
fixed  and  single  mov- 
able pulley. 


R 

too* 

FIG.  111.— Two  fixed 
and  two  movable  pul- 
leys. 


ff 

too* 

FIG.  112.— Three 
fixed  and  two  movable 
pulleys. 


jP2  are  50  Ib.  each  and  the  m.a.  is  100/50  or  2.  E  will  have 
twice  the  speed  of  R  and  the  v.r.  will  be  2. 

Figure  111  represents  two  fixed  and  two  movable  pulleys. 
The  supporting  tensions  TI,  T2,  Ts  and  T±  are  25  Ib.  each 
and  the  m.a.  is  100/25  or  4.  The  v.r.  will  also  be  4. 

Figure  112  represents  three  fixed  and  two  movable  pulleys. 
The  supporting  tensions  TI,  T2,  T3,  T4  and  T5  are  20  Ib. 
each  and  the  m.a.  is  100/20  or  5.  The  v.r.  will  also  be  5. 

From  the  above  illustrations  it  is  evident  that  the  effort 
required  to  lift  a  load  by  means  of  frictionless  pulleys  will  be 


SIMPLE  MACHINES 


119 


equal  to  the  resisting  load  divided  by  the  number  of  supporting 
strands. 

Eff          I'ft  =  resisting  load  _  _  R_ 

number  of  supporting  strands     N 

Questions  and  Problems  Relating  to  the  Pulley 

1.  Define  a  pulley.     What  is  a  block  and  tackle?     How  are  com- 
mercial pulleys  arranged  in  the  sheath?     Upon  what  principle  does  the 
pulley  operate? 

2.  For  what  purpose  is  a  single  fixed  pulley  used?     Single  movable 
pulley? 

3.  What  is  the  relation  between  the  effort,  resisting  load  and  number  of 
supporting  strands?     Show  that  this  is  a  special  application  of  the  general 
law  of  machines. 

4.  A  single  fixed  pulley  lifts  a  load  of  200  Ib.     What  effort  will  be 
required?     What  is  the  m.af     The  v.rf 

6.  A  single  movable  pulley  bears  a  load  of  300  Ib.     What  effort  will  be 
required?     What  is  the  m.a?     The  v.rf 

6.  Design  systems  of  pulleys  with  mechanical  advantages  of  1,  2,  4 
and  5. 

7.  A  1,000  Ib.  casting  is  to  be  loaded  into  a  truck.     Design  a  system  of 
pulleys  to  do  the  work,  assuming  that  the  rope  will  break  at  any  tension 
over  150  Ib. 

8.  An  "I"  beam  is  being  dragged  along  the  ground  by  means  of  a 
single  fixed  and  single  movable  pulley.     The  fixed  pulley  is  attached  to  a 
tree  and  the  movable  pulley  to  the  beam.     If  three  men  are  pulling  with 
a  force  of  50  Ib.  each,  what  force  is  exerted  (a)  on  the  beam?     (6)  On  the 
tree?     What  is  the  tension  in  each  strand? 


140.  The  Wheel  and  Axle.—  The 
wheel  and  axle  (Fig.  113)  works  on 
the  principle  of  a  first  class  lever. 
The  effort  (E,  applied  tangentially  to 


the  rim  of  the  wheel,  will  produce  a 
rotation  of  both  wheel  and  axle.  E 
will  descend  and  R  will  rise.  Repre- 
senting the  radius  of  the  wheel  by 
ri  and  the  radius  of  the  axle  by  r2, 
it  is  evident  that  E  X  <l^r\  —  R  X  27rr2,  and  that 


AR.qii 


FIG.   113.— Wheel  and  axle. 


E  = 


120 


MECHANICS 


In  Fig.  113,  r*i  is  9  in.  and  r2  is  3  in.     Hence  the  effort 

3 

necessary  to  lift  the  load  of  300  Ib.  is  300  X  g  or  100  Ib.     The 
m.a.  is  3  and  the  v.r.  is  3. 


Questions  and  Problems  Relating  to  the  Wheel  and  Axle 

1.  Describe  the  wheel  and  axle  and  state  the  purpose  for  which  it  is 
used. 

2.  State  the  mathematical  relation  of  the  effort,  resistance,  radius  of 
the  wheel  and  radius  of  the  axle. 

3.  Show  that  the  above  relation  is  an  application  of  the  general  law 
of  machines. 

4.  The  diameters  of  a  wheel  and  axle  are  24  in.  and  4  in.  respectively. 
What  load  may  be  lifted  by  an  applied  force  of  50  Ib.  ? 

6.  Compute  the  output  and  input  in  problem  4,  assuming  a  rotation  of 
720  degrees.  Also  find  the  m.a.  and  v.r. 

6.  The  m.a.  of  a  wheel  and  axle  is  10  and  the  load  to  be  raised  is  200 
Ib.  If  the  diameter  of  the  axle  is  5  in.,  what  is  the  diameter  of  the 
wheel? 

141.  The  Inclined  Plane. 

The  inclined  plane  (Fig.  1 14) 
is  a  block  whose  upper  sur- 
face makes  an  angle  of  less 
than  90°  with  the  horizontal. 
It  is  used  to  raise  heavy 
loads,  such  as  rolling  a  bar- 
rel of  sugar  up  into  a  door- 
way by  means  of  a  plank. 

The  effort  required  will  depend  on  the  slope  of  the  plane. 

According  to  the  general  law  of  machines,  the  effort  (E)  X 

the  length  (I)  of  the  plane  =  the  resistance  (#)  X  the  height 

(h)  of  the  plane. 

E  X  I  =  R  X  h  or  E  =  R  X  y  =  R  X  sin  6. 

In  the  accompanying  diagram,  the  plane  is  inclined  at  an 
angle  of  30°  to  the  horizontal  and  is  10  ft.  long  and  5  ft.  high. 
An  effort  of  50  Ib.  will  be  necessary  to  move  the  weight  of 
100  Ib.  up  the  plane.  The  m.a.  is  2  and  the  v.r.  is  2, 


FIG.   114. — Inclined  plane. 


SIMPLE  MACHINES 


121 


Questions  and  Problems  Relating  to  the  Inclined  Plane 

1.  What  is  an  inclined  plane?     For  what  purpose  is  it  used? 

2.  What  relation  exists  between  E,  R,  h  and  Z?     Between  E,  R  and 
sin  0?     Show  how  these  relations  are  derived. 

3.  An  inclined  plane  is  20  ft.  long  and  2  ft.  high.     How  heavy  a  load 
can  be  lifted  by  a  force  of  100  Ib.  applied  parallel  to  the  plane?     How 
much  work  is  done  upon  the  load  and  how  much  work  is  done  by  the 
force  if  the  load  travels  the  length  of  the  plane? 

4.  A  plane  is  inclined  at  an  angle  of  45°.     What  weight  may  be  lifted 
by  a  force  of  50  Ib.  acting  parallel  to  the  plane? 

5.  An  automobile  weighing  3,000  Ib.  is  being  towed  up  a  2  per  cent, 
grade.     Neglecting  friction,  what  is  the  tension  of  the  tow  rope? 

142.  The  Screw.— The  jack  screw  (Fig.  115)  is  a  modified 
inclined  plane.  A  piece  of  paper 
cut  into  the  shape  of  an  inclined 
plane  and  wound  around  a  pencil 
will  represent  the  threads.  The 
jack  screw  is  used  for  such  pur- 
poses as  raising  buildings  and  will 
not  reverse  when  the  applied 
force  is  removed.  Assuming  a 
complete  rotation  of  the  hand 
lever,  it  follows  that  the  effort 
(E)  X  2irr  =  the  resistance  (R) 
X  the  lead  of  the  screw.  The 
lead  is  the  distance  the  screw 
advances  during  one  rotation.  For  single-thread  screws  it  is 
the  same  as  the  pitch  (distance  between  two  adjacent  threads). 

E  X  2irr  =  R  X  lead  of  screw 

In  Fig.  115,  the  radius1  of  the  hand  lever  is  2  ft.  and  the 
lead  is  %  in.     If  the  load  (including  the  moving  part  of  the 
jack)  is  1,000  Ib.,  the  effort  to  lift  is  found  as  follows: 
R  X  lead       1,000  X 


FIG.   115. — Jack  screw. 


27TT 


2  X  TT  X  24 


=  3.31  Ib. 


Radius  of  the  hand  lever  is  measured  from  the  center  of  the  jack  to 
the  point  where  the  effort  is  applied. 


122  MECHANICS 

Questions  and  Problems  Relating  to  the  Screw 

1.  Describe  the  jack  screw.     For  what  purpose  is  it  used? 

2.  Why  does  the  jack  screw  not  reverse? 

3.  What  is  meant  by  the  lead  of  a  screw?     Pitch? 

4.  A  jack  screw  has  4  threads  to  the  inch.     If  the  radius  of  the  hand 
lever  is  3  ft.,  what  load  may  be  lifted  by  a  force  of  75  Ib.  applied  at  the 
extremity  of  the  lever?     What  is  the  ra.a?     The  v.r? 

5.  A  small  building  weighing  8,000  Ib.  is  being  raised  by  means  of  4 
jacks,  one  at  each  corner.     If  the  radius  of  the  hand  lever  is  24  in.  and  the 
pitch  of  the  threads  is  }£  in.,  what  effort  is  necessary  at  the  extremity  of 
each  lever.     What  is  the  ra.a.  and  v.r.  of  each  jack? 

143.  The  Wedge.— The  wedge  (Fig.  116)  is  simply  two 
inclined  planes  laid  base  to  base.  It  is 
used  in  splitting  timbers  and  rocks,  launch- 
ing ships,  raising  large  weights  through 
short  distances,  etc.  All  cutting  tools  are 

FIG.  116.— Wedge.       wedges. 

In  actual  practice,  friction  is  such  an 
uncertain  quantity,  that  any  law  in  regard  to  the  wedge  is 
only  a  poor  approximation. 

Questions  and  Problems  Relating  to  the  Wedge 

1.  Describe  the  wedge.     For  what  purpose  is  it  used? 

2.  In  splitting  a  piece  of  very  hard  timber  would  you  use  a  wedge  of 
large  or  small  angle?     Explain. 

3.  In  actual  practice,  the  law  of  the  wedge  is  only  approximate. 
Why? 


CHAPTER  XIX 
PRACTICAL  STUDY  OF  MACHINES 

144.  Efficiency. — It  is  assumed  that  the  student  has 
acquired,  by  this  time,  a  satisfactory  knowledge  of  the 
so-called  simple  machines.  Hereafter,  machines  will  be  dis- 
cussed from  a  practical  point  of  view.  It  is  impossible,  in  this 


FIG.  117. — Comparison  of  a  Yale  differential,  screw-geared  and  spur-geared 

chain  block. 

small  volume,  to  examine  many  of  the  machines  in  common 
use.  Those  selected,  however,  will  be  typical  and  furnish  an 
excellent  background  for  further  study. 

In  Chap.  18,  the  general  law  of  machines  is  stated: 

E  X  di  X  Effic.  =  R  Xd2 
123 


124  MECHANICS 

Since  d\/di  is  equal  to  the  velocity  ratio,  the  law  may  also 
be  stated: 

E  X  V.R.  X  Effic.  =  R  X  1 

Thus,  to  find  the  input  of  a  machine,  determine  the  velocity 
ratio  and  multiply  it  by  the  effort.  The  output  will  be  equal 
to  the  resistance  times  1  or,  simply,  the  load.  The  above 
method  gives  the  relative  input  and  output  and  is  independent 
of  the  actual  distance  covered  by  E  and  R.  It  is  suggested 
that  all  problems  be  solved  according  to  this  method. 

145.  Chain  Blocks. — As  a  hoisting  device,  the  chain  block  is 
universally   used.     It    is    almost   indispensable    in    machine 
shops,  factories,  warehouses,   etc.,   where  heavy  loads  must 
be  lifted  frequently.     There  are  three  general  types  in  common 
use:  the  differential  block;  the  screw-geared  block;  and  the 
spur-geared  block. 

Figure  117  shows  the  Yale  differential,  screw-geared  and 
spur-geared  chain  blocks  of  one  ton  capacity.  The  picture 
shows  the  maximum  load  that  each  man  can  comfortably 
lift  and  the  distance  through  which  the  load  can  be  lifted  in 
30  sec.  It  is  assumed  that  each  man  pulls  82  Ib.  The 
student  will  observe  that  the  spur-geared  block  has  an  advan- 
tage both  of  speed  and  efficiency  over  the  other  two. 

146.  Differential     Chain    Block.— The    differential    chain 
block  employs  two  sheaves  of  slightly  different  diameter  in 
the  upper  block  and  one  sheave  in  the  lower  block.     The 
sheaves  in  the  upper  block  are  cast  together.     Over  these 
sheaves  runs  an  endless  hand  chain.     Pockets  along  the  rims 
of  the  upper  sheaves  prevent  slippage.     When  the  effort  is 
removed,  the  friction  of  the  blocks  is  sufficient  to  restrain  the 
load  from  running  back.     The  differential  is  simple,  cheap  and 
reliable,  but  has  not  the  durability  or  efficiency  of  the  screw- 
geared  or  spur-geared  block. 

The  velocity  ratio  of  the  differential  is  determined  as  follows.  • 
Referring  to  Fig.  .118  and  assuming  that  the  upper  sheaves 
make  one  revolution,  E  then  moves  a  distance  of  2irR.     Side 
B  of  the  chain  loop  is  raised  2irR,  but  side  A  is  lowered  2irr 


PRACTICAL  STUDY  OF  MACHINES 


125 


as  the  chain  unwinds.  The  loop  A B  is  therefore  shortened 
2irR  —  2irr  or  2ir(R  —  r).  The  load  is  raised  only  half  this 
distance  or  x  (R  —  r).  Thus  the  effort  distance  divided  by  the 
load  distance  equals  2R/R  —  r. 

Velocity  ratio  for  differential  = 


—  r 
Since  R  and  r  are  proportional  respectively  to  the  number 

of  chain  pockets  in  the  larger  and  smaller  sheaves  of  the  upper 

block,  it  is  suggested  that  these  values  be 

substituted  for   R   and    r  in  the   above 

equation.     This   method   eliminates  any 

error  in  measurement. 

In  conducting  a  laboratory  study  of  the 

differential,    the    velocity    ratio    is    first 

determined.     This   is  done  by   counting 

the  chain  pockets  in  the  upper  sheaves  and 

substituting  these  values  in  the  formula 

given    above.      The   calculated    velocity 

ratio  is  checked  by  finding  the  relative 

distances  moved  by  the  effort  and  load. 
Next  it  determined,  by  means  of  a  spring 

balance,  the  effort  necessary  to  run  the 

machine  unloaded.     A  load  of  50  Ib.  is 

now  suspended  from  the  load  hook  and  the 

effort  to  lift  recorded.  The  load  is  in- 
creased 50  Ib.  at  a  time  and  the  corre- 
sponding efforts  to  lift  taken  in  each  case. 

'    The  following  figures  were  submitted  by  two  students.     The 

chain  block  (quarter-ton  capacity)  was  manufactured  by  the 

Chisholm  and  Moore  Mfg.  Co.,  Cleveland,  Ohio. 

VELociTy  RATIO 

(a)  Chain  pockets  in  larger  pulley  of  upper  block  =  9. 
(6)  Chain  pockets  in  smaller  pulley  of  upper  block  =  8. 

O  S/  Q 

(c)  Computed  velocity  ratio  =  ^      „  =  18. 


FIG.  i is.— Differential 


9- 
(d)  Actual  velocity  ratio  (check), 

Distance  effort  travels  _  108  in. 
Distance  load  travels          6  in. 


•=  18, 


126 


MECHANICS 
EFFICIENCY 


Load  in 

Effort  to 

Velocity 

Input  in 

Output  in 

Efficiency 

Ib. 

lift  in  Ib. 

ratio 

ft.  Ib. 

ft.  Ib. 

in  per  cent. 

0    . 

2 

18 

36 

. 
0 

0 

50 

11 

18 

198 

50 

25.3 

100 

18 

18 

324 

100 

30.8 

150 

26 

18 

468 

150 

32.0 

200 

34 

18 

612 

200 

32.6 

250 

43 

18 

774 

250 

32.3 

300 

51 

18 

918               300 

32.6 

150  200 

Load  in  Pounds 


250 


300 


FIG.  119. — Graph  showing  relation  of  efficiency  and  lifting  effort  to  load 
for  the  differential  chain  block. 

Figure  119  is  a  graph  showing  the  relation  of  the  driving 
effort  and  efficiency  to  the  load.  It  is  customary  to  locate  the 
loads  along  the  "x"  or  horizontal  axis  and  the  efforts  and 
efficiencies  along  the  "y"  or  vertical  axis.  This  graph  is 
typical  for  machines  and  should  be  thoroughly  understood  by 
the  student.  The  load-effort  curve  starts  above  zero,  as  some 
effort  is  required  to  drive  the  hoist  unloaded.  The  efficiency 
curve  always  has  its  origin  at  zero,  since  the  efficiency  at  zero 
load  must  necessarily  be  zero.  It  will  be  observed  that  the 


PRACTICAL  STUDY  OF  MACHINES 


127 


effort  is  directly  proportional  to  the  load  and  that  the  efficiency 
rises  with  the  load  to  about  32  per  cent,  and  then  remains  prac- 
tically constant. 

145.  Screw-geared  Chain  Block. — The  screw-geared  block 
is  light,  compact  and  especially  adapted  to  portable  work. 
It  is  cheaper  than  the  spur-geared,  although  it  does  not  possess 
the  speed  of  the  latter.  It  depends  upon  the  worm  and  worm 
wheel  for  its  mechanical  advantage  (See  Fig.  120).  Effort 
is  applied  by  means  of  a  hand  chain  over  a  sheave  having 
pockets  for  the  chain  links.  This  sheave  is  a  unit  with  the 


FIG.    120. — Working  mechanism  of  a  Yale  screw-geared  chain  block. 

worm  shaft  which  directly  operates  the  worm  wheel.  The 
worm  wheel  in  turn  rotates  the  load  sheaves.  The  worm 
and  worm  wheel  are  of  steel  and  bronze  respectively.  Parts 
subject  to  friction  run  in  oil,  insuring  proper  lubrication.  The 
block  may  be  reversed  but  will  not  "run  down." 

To  compute  the  velocity  ratio,  the  diameter  of  the  hand 
wheel  and  the  diameter  of  the  load  sheaves  is  first  determined. 
The  diameter  (Di)  of  the  hand  wheel  is  equal  to  the  distance 
between  the  mid-points  of  the  chain  running  around  it.  The 


128 


MECHANICS 


diameter  (Dz)  of  the  load  sheaves  is  determined  in  a  similar 
manner.  Next  the  number  (N)  of  teeth  in  the  worm  wheel  is 
determined  by  counting  the  number  of  revolutions  of  A 
required  to  rotate  C  once. 

Velocity  ratio  for  screw-geared  block  =  ~  X  N. 

JL>Z 

In  the  laboratory,  the  velocity  ratio  is  found  as  described 
above  and  checked  by  noting  how  far  the  hand  chain  moves 
while  the  load  chain  moves  1  ft. 

Next  the  effort  to  drive  the  block  unloaded  is  determined 
by  a  spring  balance.  Various  loads  are  now  suspended  from 
the  load  hook  and  the  corresponding  efforts  to  lift  recorded. 

The  following  test  of  a  Yale  block  was  made  by  two  students. 

VELOCITY  RATIO 

(a)  Diameter  of  hand  wheel  (A)  =  5.25  in. 
(6)  Diameter  of  load  sheaves  (D2)  =  2.75  in. 

(c)  Number  of  revolutions  of  A  to  one  of  C  =  20. 

(d)  Calculated  velocity  ratio  =  ^  X  N  =  |^  X  20  =38.2. 

(e)  Distance  moved  by  hand  chain  =  456  in. 
(/)  Distance  moved  by  load  hook  =  12  in. 

A  f^A 

(g)  Actual  velocity  ratio  (check)  =  -^  =  38. 
EFFICIENCY 


Load  in 
Ib. 

Effort  to 
lift  in  Ib. 

Velocity 
ratio. 

Input  in 
ft.-lb. 

Output  in 
ft.-lb. 

Efficiency 
in  per  cent. 

0 

1 

38 

38 

0 

0 

55 

4 

38 

152 

55 

36.1 

105 

7 

38 

266 

105 

39.4 

155 

10 

38 

380 

155 

40.7 

205 

13 

38 

494 

205 

41.5 

255 

16                  38 

608 

255 

41.9 

Figure  121  is  a  graph  showing  the  relation  of  effort  and 
efficiency  to  the  load.  An  examination  of  the  graph  shows  that 
the  effort  is  directly  proportional  to  the  load  and  that  the  effi- 


PRACTICAL  STUDY  OF  MACHINES 


129 


ciency  rises  with  the  load,  becoming  nearly  constant  around  41 
per  cent.  Note  that  the  efficiency  curve  starts  at  zero  and  that 
the  load-effort  curve  starts  above  zero. 


50 


40 


30 


o 


40 


Z40 


80          IZO          160         200 
Load  in  Pounds 

FIG.   121. — Graph  showing  relation  of  efficiency  and  lifting  effort  to  load  for 
the  screw-geared  chain  block. 

148.  Spur -geared    Chain    Block. — The    Yale    spur-geared 
block  is  very  efficient,  frictional  losses  being  reduced  to  a 
minimum.     It  is  made  for  capacities  of  from  J4  ton  to  40 
tons.     One  man  pulling  82  Ib.  on  the  hand  chain  can  lift 
one  ton.     The  lowering  of  the  load  is  effected  by  an  automatic 
brake  mechanism.     Referring  to  Fig.  122,  it  will  be  seen  that 
the  block  employs  spur  gears  based  on  the  principle  of  the 
planetary    system.     Movement    of    the    hand    chain    rotates 
pinion  A.     The  pinion  transmits  its  energy  to  two  intermediate 
gears  (B  and  B)  diametrically  opposed.     The  intermediate 
gears  mesh  with  the  large  internal  gear  D,  causing  a  revolution 
of  the  pinion  cage  E  to  which  the  load  sheave  F  is  attached. 
The  design  of  the  block  distributes  the  pressure  equally  and 
tends  to  prevent  wear  in  the  bearings. 

149.  Chain-drive  Bicycle. — The  bicycle  is  a  speed  machine, 
effort  being  sacrificed  in  order  to  obtain  a  high  velocity  for  the 
rear  wheel.     It  is  evident  that  both  the  velocity  ratio  and 

o 


130 


MECHANICS 


FIG.  122. — Working  mechanism  of  a  Yale  spur-geared  chain  block. 

mechanical  advantage  will  be  less  than  1.     The  bicycle  in 
Fig.  123  is  intended  for  practical  study.     It  is  firmly  mounted 


FIG.  123. — Bicycle  mounted  for  laboratory  test. 

and  the  pedals  have  been  replaced  by  a  wooden  pulley  equal  in 
radius  to  the  length  of  the  pedals.     Thus  the  efforts  jnay  be 


PRACTICAL  STUDY  OF  MACHINES 


131 


applied  conveniently  by  means  of  a  stout  cord  and  a  scale  pan. 
The  loads  are  suspended  from  the  rear  wheel  in  a  similar 
manner. 

In  a  practical  test,  the  velocity  ratio  is  first  calculated.  The 
circumference  of  the  driving  and  driven  wheel  is  determined 
and  the  number  of  teeth  in  the  front  and  rear  sprockets. 

circumference  wooden  pulley 
V.r.  for  bicycle  =  —  :  --  7—  -~  —  ^  X 

circumference  rear  wheel 

teeth  rear  sprocket 
teeth  front  sprocket 

The  velocity  ratio  is  checked  by  finding  the  simultaneous 
distances  travelled  by  the  effort  and  the  load. 

Next  it  is  determined  how  much  weight  must  be  applied  to 
the  driving  pulley  to  run  the  bicycle  unloaded.  Various 
weights  are  suspended  from  the  rear  wheel  and  the  correspond- 
ing efforts  to  lift  found. 

The  figures  below  were  submitted  by  a  student  in  the 
Wm.  L.  Dickinson  H.  S. 


VELOCITY  RATIO 


(a) 


Circumference  of  driving  pulley  =  47.2  in. 

(b)  Circumference  of  rear  wheel  =  77  in. 

(c)  Teeth  on  rear  sprocket  =  8. 
(eO  Teeth  on  front  sprocket  =  24. 

(e)  Distance  effort  travels  while  load  travels  36  in.  =  7.34  in. 
(/)  Computed  velocity  ratio  =  47.2/77  X  8/24  =  .204. 
(g)  Actual  velocity  ratio  =  7.34/36  =  .204. 

EFFICIENCY 


Load  in 
Ib. 

Effort  to 
lift  in  Ib. 

Velocity 
ratio. 

Input  in 
ft.-lb. 

Output  in 
ft.-lb. 

Efficiency 
in  per  cent. 

0 

0.13 

0.204 

0.026 

0 

0 

0.25 

1.37 

0.204 

0.279 

0.25 

89.6 

0.50 

2.62 

0.204 

0.534 

0.50 

92.9 

1.00 

5.12 

0.204 

1.044 

1.00 

95.8 

2.00 

10,12 

0.204 

2.064 

2.00 

96.9 

4.00 

20.12 

0.204 

4.104 

4.00 

97.4 

6.00 

30.20 

0.204 

6.160 

6.00 

97.4 

132 


MECHANICS 


Figure  124  is  a  graph  showing  the  relation  of  effort  and 
efficiency  to  load  for  the  bicycle.  The  load-effort  line  begins 
slightly  above  zero,  indicating  that  the  amount  of  friction  is 
very  small  as  compared  with  the  machines  previously  studied 
in  this  chapter.  It  is  evident  that  the  effort  is  directly  pro- 
portional to  the  load  and  that  the  efficiency  increases  with  the 
load  to  a  little  over  97  per  cent,  and  then  remains  constant. 


(A 

-o 

c 

I 


UJ 


-EFFICIENCY- 


100 


75 

Q> 
Q- 
C 


50 


UJ 


Z5 


l  2  3_  4  5  G 

Load  in  Pounds 

FIG.  124. — Graph  showing  relation  of  efficiency  and  lifting  effort  to  load  for 

the  bicycle. 

150.  The  Jack  Screw. — The  principle  of  the  jack  screw  was 
discussed  in  Chap.  18.  In  order  to  approach  actual  working 
conditions  for  an  efficiency  test,  the  apparatus  shown  in 
Fig.  125  is  recommended.  Sufficient  loads  may  thus  be 
obtained  to  make  the  test  very  practical. 

An  examination  of  the  above  diagram  shows  that  the  hand 
lever  has  been  replaced  by  a  grooved  wheel  equal  in  radius  to 
the  length  of  the  original  torque  rod.  Various  weights  are 
suspended  from  D  and  the  corresponding  efforts  to  rotate  the 
wheel  are  determined  by  means  of  a  spring  balance  attached  to 
a  stout  cord  running  around  the  wheel.  The  experimental 
work  is  conducted  as  follows. 

First  the  velocity  ratio  is  determined.     This  is  found  by 


PRACTICAL  STUDY  OF  MACHINES 


133 


dividing  the  circumference  of  the  wheel  by  the  lead  of  the  screw. 

. .     .      .    7  circumference  of  driving  wheel 

Velocity  ratio  for  jack  screw  =  -          — -, — 3—7 — 

lead  of  screw 

Next  the  weight  and  center  of  gravity  of  the  beam  is  found ; 
also  the  weight  of  the  moving  part  of  the  jack.  It  is  sug- 
gested that  these  figures  be  recorded  permanently  on  the  beam 

B  C 


FIG.   125. — Laboratory  apparatus  for  test  of  a  jack  screw. 


r 


< 4 


t  * i 


W  W, 

FIG.   126. — Force  diagram  for  the  jack  screw. 

for  quick  reference  and  that  yard  sticks  be  attached  to  the 
beam  to  facilitate  the  measurement  of  distances.  The  beam 
is  levelled  and  50  Ib.  is  suspended  from  D.  The  effort  to 
rotate  the  wheel  slowly  is  found  by  means  of  a  spring  balance. 
The  suspended  weight  is  increased  50  Ib.  at  a  time  and  the 
efforts  to  lift  recorded  in  each  case;  also  the  distances  di,  di 
and  dz  (See  Fig.  126).  The  load  may  also  be  varied  by  moving 
the  jack  along  the  beam.  This  is  not  so  convenient,  however. 
To  find  the  total  load  (L2)  raised  by  the  effort  (E),  the 
following  procedure  is  used.  Referring  to  Figs.  125  and  126, 


134 


MECHANICS 


it  will  be  seen  that  A  in  the  picture  diagram  becomes  F  in  the 
force  diagram  and  is  considered  the  fulcrum  of  the  lever. 
The  "down"  forces  are  the  weight  (W)  of  the  lever  concen- 
trated at  the  center  of  gravity  of  the  beam  and  the  suspended 
weight  (Wi)1.  The  "up"  force  is  the  pressure  exerted  against 
the  beam  at  B  and  is  labelled  L  on  the  force  diagram. 

Referring  to  Fig.  126  and  applying  a  suitable  law  of  levers, 
it  will  be  seen  that, 

L  X  di  =  W  X  d2  +  W i  X  (d2  +  d8) 
L  is  determined  in  each  case  by  the  above  formula.     The 

total  load  (L2)  is  equal  to  L  plus  the  weight  of  the  moving 

part  of  the  jack. 

The  following  experimental  work  on  the  jack  screw  was  done 

in  the  mechanics  laboratory  of  the  Dickinson  High  School. 

VELOCITY  RATIO 

(a)  Circumference  of  hand  wheel  =  59  in. 

(6)  Lead  of  screw  =  .50  in. 

(c)  Computed  velocity  ratio  59/.50  =  1.18. 

DATA 


Weight   of 

Weight  of 

Suspended 

Effort 

beam  (W)  in 

jack  in  Ib. 

weight  (Wi) 

(E)   in 

di  in  in. 

di  in  in. 

di  in  in. 

Ib. 

in   Ib. 

Ib. 

1 

23 

36 

50 

18 

12 

44.5 

49 

2 

23 

36 

100 

31 

12 

44.5 

49 

3 

23 

36 

150 

44 

12 

44.5 

49 

EFFICIENCY 


Total  load  (L2) 

Input  in  ft.-lb. 

Output  in  ft.- 

Efficiency  in 

inlb. 

lb. 

per  cent. 

1 

511 

2,124 

511 

24.0 

2 

900 

3,658 

900 

24.6 

3 

1,290 

5,192 

1,290 

24.8 

The  "down"  force  at  F  is  not  considered,  as  its  moment  is  zero. 


PRACTICAL  STUDY  OF  MACHINES 


135 


Figure  127  is  a  graph  constructed  from  the  above  figures. 
The  low  efficiency  is  accounted  for  by  the  excessive  friction. 
As  in  the  previous  machines,  the  effort  is  directly  proportional 
to  the  load.  The  efficiency  remains  constant  at  around  24  per 
cent.  The  part  of  the  efficiency  curve  where  no  figures  were 
taken  is  represented  by  a  broken  line. 


CTA 

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X 

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a 

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fjv 

/ 

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1C         C 

/ 

'/rf* 

^ 

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b   5^ 

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/ 

/ 

^ 

^x 

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IU  jj- 
ua 

in 

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20 

0 

4C 

0 

feC 

0 

SC 

0 

10 

30. 

I?C 

0 

I4C 

0 

Load  in  Pounds 

FIG.   127. — Graph  showing  relation  of  efficiency  and  lifting  effort  to  load  for 

the  jack  screw. 

151.  The  Automobile  Transmission. — Since  a  gasoline 
motor  car  delivers  its  maximum  power  at  high  speeds,  it  is 
absolutely  essential  that  some  device  be  provided  for  varying 
the  velocity  ratio  between  the  crankshaft  and  the  rear  axles. 
This  is  effected  by  a  special  set  of  spur  gears  called  the  trans- 
mission or  change  gears. 

Figure  128  shows  a  typical  transmission.  It  consists  of  a 
main  shaft  and  countershaft,  each  carrying  spur  gears  and 
mounted  on  roller  bearings. 

The  main  shaft  carries  three  gears:  (1)  the  main  drive  gear 
which  receives  the  power  from  the  motor;  (2)  the  high  and 
intermediate  sliding  gear;  and  (3)  the  low  and  reverse  sliding 
gear.  The  sliding  gears  are  arranged  to  move  along  their 
shaft  by  moans  of  a  shifter  fork,  but  the  shaft  and  gears  must 


136 


MECHANICS 


rotate  together.  A  shaft  along  which  gears  can  slide  and 
which  rotates  as  a  unit  with  the  gears  is  often  called  a  spline 
shaft. 

The  countershaft  carries  four  gears,  each  solid  with  the 
shaft:  (1)  the  countershaft  drive  gear;  (2)  the  countershaft 
second  speed  gear;  (3)  the  countershaft  first  speed  gear;  and  (4) 
the  countershaft  reverse  gear. 

The  neutral  position  of  the  gears  in  Fig.   128  allows  the 

.-Low  and  Reverse 

nic/n  and  Intermediate       y>  Sliding  &ear 
Sliding  Gear 

Main  Drive  Gear: 


ToEnginel     r 


Main  Drive  Shaft 


To  Final 
"  Drive 


Counter-,. 

Shaff 

Drive 

Gear 


'"Reverse  Gear 


FIG.   128.  —  Typical  automobile  transmission. 


motor  to  run  without  transmitting  any  power  to  the  propeller 
shaft.  Since  there  is  a  "  break"  between  the  main  drive  gear 
and  the  high  and  intermediate  sliding  gear,  the  countershaft 
merely  idles  and  the  part  of  the  main  drive  shaft  connected 
to  the  final  drive  and  carrying  the  sliding  gears  is  cut  off  from 
the  source  of  power. 

First  speed  or  "low,"  used  in  starting  and  for  heavy  loads, 
provides  the  highest  velocity  ratio  of  the  forward  speeds.  The 
low  and  reverse  sliding  gear  is  moved  to  the  left  by  the 
shifter  fork,  engaging  with  the  first  speed  gear  on  the  counter- 


PRACTICAL  STUDY  OF  MACHINES  137 

shaft.  Thus  the  propeller  shaft  receives  its  motion  from  the 
motor  by  way  of  the  countershaft. 

Second  speed  or  "intermediate"  is  obtained  by  moving  the 
high  and  intermediate  sliding  gear  to  the  right,  so  that  it  will 
engage  with  the  second  speed  gear  on  the  countershaft.  The 
drive  is  through  the  countershaft  and  the  velocity  ratio  is 
less  than  in  the  preceding  case. 

For  third  speed  or  "high,"  the  high  and  intermediate  sliding 
gear  is  moved  to  the  left,  it  internal  teeth  meshing  with  the 
external  teeth  at  the  right  of  the  main  drive  gear.  Thus,  in 
third  speed,  the  upper  shaft  rotates  as  a  unit,  while  the  counter- 
shaft idles.  It  will  be  seen  that  the  motor  crankshaft  and  the 
propeller  shaft  have  the  same  number  of  r.p.m.  The  velocity 
ratio  is  less  than  in  second  speed. 

For  reverse  speed,  the  low  and  reverse  sliding  gear  is  moved 
to  the  right,  engaging  with  the  reverse  idler.  The  reverse 
idler  is  always  in  mesh  with  the  reverse  gear  on  the  counter- 
shaft and  runs  idle  except  in  reverse  speed.  The  drive  is 
through  the  countershaft. 

It  should  be  noted  that  the  two  sliding  gears  are  never  both 
in  mesh  at  the  same  time.  The  shifter  fork  always  removes 
the  gear  in  mesh  before  putting  the  other  gear  in  engagement. 

Questions  and  Problems 

1.  State  a  general  law  of  machines  involving  effort,  efficiency,  load  and 
velocity  ratio. 

2.  Discuss  the  construction   and  advantages  of  the  chain   blocks 
described  in  this  chapter. 

3.  State  how  you  would  compute  the  velocity  ratio  of  the  above 
blocks. 

4.  Discuss  the  bicycle  as  a  machine.     How  is  the  velocity  ratio 
computed? 

5.  Describe  a  practical  laboratory  test  for  the  jack  screw.     How  is 
the  velocity  ratio  calculated? 

6.  What  general  conclusions  did  you  draw  from  a  study  of  the  graphs 
in  this  chapter?     Why  does  an  efficiency  curve  always  start  at  the 
intersection  of  the  "x"  and  "y"  axis?     Why  does  the  load-effort  graph 
always  start  above  the  intersection? 

7.  What  is  the  purpose  of  the  automobile  transmission?     Describe 


138 


MECHANICS 


the  transmission  in  detail,  diagramming  the  position  of  the  gears  for 
each  speed. 

8.  A  differential  type  of  chain  hoist  has  8  and  7  chain  pockets  respec- 
tively in  the  sheaves  of  the  upper  block.     What  is  the  velocity  ratio? 
If  the  efficiency  is  30  per  cent.,  how  much  effort  is  needed  to  lift  a  load 
of  100  lb.? 

9.  The  velocity  ratio  of  a  screw-geared  chain  block  is  38.     If  the 
efficiency  is  40  per  cent.,  what  load  can  be  raised  by  an  applied  force  of 
10  lb? 

10.  In  a  laboratory  test  of  a  chain-drive  bicycle  (see  Fig.  123),  a  force 
of  25  lb.  applied  to  the  driving  pulley  was  necessary  to  overcome  5  lb. 
applied  to  the  rear   wheel.     What   is    the 
velocity  ratio  if  the  efficiency  at  this  load 
is  97  per  cent.  ? 

11.  The  velocity  ratio  of  a    jack  screw 
arranged  as  in  Fig.    125    is    200.      If    the 
efficiency  is  25  per   cent.,    what   effort    is 
"^••-^"O        necessary  to  raise  a  total  load  of  l,5001b.? 


FIG.   120. 


FIG.  130. 


12.  A  wheel  12  in.  in  diameter  makes  1,000  r.p.m.     It  is  belted  to  a 
pulley  6  in.  in  diameter.     The  pulley  is  keyed  to  a  second  wheel  30  in.  in 
diameter.     Assuming  a  belt  slippage  of  10  per  cent.,  determine  the  linear 
speed  of  a  point  on  the  circumference  of  the  second  wheel. 

13.  The  screw  of  the  bench  vise  in  Fig.  129  has  9  threads  to  the  inch. 
The  effective  lever  arm  of  the  hand  lever  is  5  in.     Allowing  a  65  per  cent, 
loss  due  to  friction,  what  clamping  force  is  exerted  by  the  jaws,  if  the 
applied  effort  is  15  lb.  ? 

14.  The  turnbuckle  in  Fig.   130  has  12  threads  to  the  inch.     The 
threaded  ends  of  the  rods  are  2  in.  apart.     How  far  apart  will  they  be 
after  10  complete  turns?     Give  two  answers. 

15.  Devise  a  system  of  pulleys  whereby  one  man  pulling  not  over  75 
lb.  can  lift  a  weight  of  300  lb.     Assume  an  efficiency  of  75  per  cent. 
Label  the  diagram  carefully  and  prove  its  correctness. 

16.  The  screw  of  a  %  in.  bolt  has  12  threads  to  the  inch.     The  effective 
leverage  of  the  wrench  used  in  turning  on  the  nut  is  8  in.     What  is  the 
clamping  force  exerted  by  the  nut,  if  the  effort. applied  to  the  wrench  is 
10  lb.  ?     Assume  a  frictional  loss  of  70  per  cent. 

17.  A  machine  delivering  8  hp.  loses  500  ft.-lb.  of  work  per  second 
due  to  friction.    What  is  the  efficiency  of  the  machine? 


PRACTICAL  STUDY  OF  MACHINES  139 

18.  A  simple  hoisting  winch  consists  of  a  pinion  gear  turned  by  a 
crank  and  engaging  a  larger  gear.     The  larger  gear  turns  a  barrel  around 
which  the  load  chain  winds.     Crank  lever  =  20  in.;  teeth  in  pinion 
=  24;  teeth  in  larger  gear  =  72;  diameter  of  barrel  =  14  in.;  load  = 
200  lb.;  efficiency  =  75  per  cent.     Find  (a)  the  velocity  ratio  and  (b) 
the  effort  to  lift,  if  the  force  is  applied  perpendicular  to  the  lever. 

19.  An  automobile  weighing  3,000  lb.  is  being  towed  up  a  2  per  cent, 
grade  (2  ft.  rise  in  a  hundred  ft.).     Assuming  friction  to  be  15  per  cent, 
of  the  weight,  what  is  the  tension  in  the  tow  rope? 

20.  An  electric  motor,  with  a  driving  pulley  6  in.  in  diameter,  makes 
1,500  r.p.m.     This  pulley  is  belted  to  another  pulley  8  in.  in  diameter. 
The  latter  pulley  is  keyed  to  a  shaft  terminating  in  bevel  pinion  with 
15  teeth  and  engaging  a  bevel  gear  with  75  teeth.     The  bevel  gear  turns 
a  third  pulley  10  in.  in  diameter.     How  many  r.p.m.  does  the  10  in. 
pulley  make?     Assume  no  belt  slippage. 


21.  The  "whip  on  whip"  pulley  arrangement  shown  in  Fig.   131  is 
used  for  dragging  guns  out  of  ditches,  etc.     B  is  fastened  to  a  convenient 
tree,  let  us  say;  R  to  the  gun  mount;  and  the  snub  to  a  stump.     The 
effort  is  applied  at  F.     Assuming  an  efficiency  of   100  per  cent,  and 
that  the  ropes  and  blocks  have  no  weight,  find:  (a)  the  relative  distances 
travelled  by  F,  A  and  R;  (6)  the  tension  in  each  strand,  if  8  soldiers  are 
exerting  a  force  of  500  lb.  at  F;  (c)  the  reaction  at  B. 

22.  In  studying  the  transmission  of  a  Studebaker  six  cylinder  auto- 
mobile  (gears  as  in  Fig.    128),   two  students  submitted  the  following 
data:  teeth  on  ring  gear  driving  axles  =  52;  teeth  on  pinion  driving 
ring  gear  =  14;  teeth  on  main  drive  gear  =  16;  teeth  on  low  and  reverse 
sliding  gear  =31;  external  teeth  on  high  and  intermediate  sliding  gear 

=  24;  teeth  on  countershaft  gears  from  left  to  right  =  32,  24,  17  and  13 
respectively.  Compute  the  velocity  ratio  between  the  motor  crank- 
shaft and  the  rear  axles  for  (a)  first  speed;  (6)  second  speed;  (c)  third 
speed;  (d)  reverse  speed. 


CHAPTER  XX 
MECHANICAL  TRANSMISSION  OF  POWER 

152.  Transmission  of  Power. — Various  methods  are 
employed  in  transmitting  power  from  one  point  to  another. 
The  kind  of  transmission  used  will  depend  upon  the  nature  of 
the  work  to  be  done.  Mechanical  transmission  is  effected 
by  means  of  the  following  devices: 

1.  Shafts 

2.  Couplings 

3.  Clutches 

4.  Cams 

5.  Links 

6.  Screw  Threads 

7.  Chains  and  Sprockets 

(a)  Pulleys  and  Belts 


8.  Friction  Devices 


(b)  Pulleys  and  Ropes 


(c)  Friction  wheels 

9.  Toothed  Gears 

153.  Shafts. — A  shaft  is  a  rotating  bar  used  in  transmitting 
power.  Shafts  are  usually  of  steel.  Alloy  steel,  such  as 
nickel  or  vanadium  steel,  is  used  for  heavy  work  and  when 
weight  is  to  be  kept  down.  Marine  engine  and  automobile 
shafts  are  generally  alloys.  Hollow  shafts  have  found  much 
favor,  as  they  are  stronger  in  proportion  to  their  weight  than 
solid  shafts. 

A  line  shaft  (Fig.  132)  is  made  up  of  several  lengths  of 
shafting  coupled  together  and  forming  a  continuous  run.  It 
may  or  may  not  be  a  main  line  shaft. 

The  main  shaft  is  a  line  of  shafting  attached  directly  to  the 
prime  mover  (motor  or  engine).  The  crankshaft  of  an  auto- 
mobile is  a  part  of  the  main  drive  shaft  (Fig.  133). 

140 


MECHANICAL  TRANSMISSION  OF  POWER  141 


FIG.  132. — Line  shaft  used  in  a  beet  sugar  factory. 


VALVES 


SHAFT 


CRANK  SHAFT 


FIG.  133. — Working  parts  of  a  Ford  motor. 


FIG.  134. — Countershaft  used  in  shops  and  factories. 


142 


MECHANICS 


The  countershaft  is  located  between  the  main  shaft  and  the 
driven  machine.  It  is  usually  a  short  section  and  serves 
principally  to  effect  changes  of  speed  and  direction.  The 
cam  shaft  of  an  automobile  (Fig.  133)  is  an  example.  The 
Dodge  countershaft  shown  in  Fig.  134  is  widely  used  in  fac- 
tories and  shops.  Attention  is  also  directed  to  the  trans- 
mission countershaft  in  Fig.  128,  by  means  of  which  various 
velocity  ratios  are  made  possible  between  the  motor  crank- 
shaft and  the  rear  wheels. 

A  secondary  shaft  is  here  used  to  include  all  shafts  except 
main  and  countershafts.  The  spline  shaft  in  Fig.  128,  is  a 
secondary  shaft  except  in  third  speed.  In  third  speed,  it  is  a 
part  of  the  main  shaft. 

A  flexible  shaft  is  used  in  transmitting  power  at  a  curved 
distance  from  the  source.  The  dentist's  drill  and  boring  tools 
for  mines,  etc.,  are  examples. 

154.  Couplings. — Line  shafts  exceeding  25  ft.  in  length  are 
usually  made  up  of  separate  sections  joined  together  by 


FIG.  135.— Flange  fixed  coupling. 


FIG.  136. — Solid-sleeve  fixed 
coupling. 


couplings.     Couplings  are  also  used  for  short  sections  as  in  a 
motor-generator  set. 

Fixed  couplings  are  used  when  a  shaft  is  subject  to  rotation 
only  and  is  free  from  sudden  and  dangerous  torques.  The 
flange  type  of  fixed  coupling  (Fig.  135)  is  generally  used  on 
large  shafts.  The  solid  sleeve  fixed  coupling  (Fig.  136)  is 


MECHANICAL  TRANSMISSION  OF  POWER 


143 


suitable  for  small  shafts.     Other  types  of  fixed  couplings  are 
omitted  through  lack  of  space. 

Flexible  couplings  are  used  for  shafts  liable  to  poor  alignment 
or  subject  to  sudden  and  dangerous  torques.  The  flange 
flexible  coupling  (Fig.  137)  is  used  for  direct  connections,  such 


FIG.  137.— Flange  flexible 
coupling. 


FIG.  138.- 


-Flexible  coupling  of  an  automobile 
starting  motor. 


as  gas  engines,  turbines,  motors,  generators,  etc.  The  opposite 
parts  are  held  together  by  flexible  pins  instead  of  rigid  bolts. 
The  pins  are  made  of  tempered  steel  leaves.  Figure  138 
represents  an  Auto-Lite  starting  motor  used  on  Chevrolet 
motor  cars.  The  driving  pinion,  engaging  with  teeth  on  the 
flywheel,  is  connected  to  the  armature  shaft  by  means  of  a 
coiled  spring.  Thus  the  spring  absorbs  much  of  the  sudden 
initial  torque. 

155.  Universal   Couplings. — Universal   couplings   are   used 


FIG.  139. — Single  universal 
joint. 


FIG.  140. — Double  universal  joint. 


for  shafts  subject  to  disalignment,  as  the  propeller  shafts  of 
automobiles,  etc. 

The  universal  couplings  shown  in  Figs.  139  and  140  are 
constructed  with  steel  bolts  fitting  into  bronze  bushings.    The 


144 


MECHANICS 


universal  coupling  is  low  in  efficiency,  but  absolutely  neces- 
sary for  certain  work. 

156.  Clutches.— Couplings  which  admit  of  ready  disengage- 
ment are  called  clutches.     Clutches  are  usually  of  the  positive 


FIG.  141. — Common  types  of  jaw  clutches. 

type  or  the  friction  type,  although  magnetic  clutches  have 
been  used  to  some  extent. 

The  jaw  clutches  shown  above  are  of  the  positive  type. 
It  will  be  noted  that  the  opposite  parts  interlock,  eliminat- 
ing any  possibility  of  slippage. 

The  friction  clutch  utilizes  fric- 
tion to  hold  the  opposite  parts 
together.  Friction  clutches  are 
generally  of  the  cone  or  the  disc 
type,  the  contracting-band  and 
expanding-band  clutches  having 
become  nearly  obsolete. 

Figure  142  shows  a  cone  clutch 
such  as  is  used  on    automobiles. 
Friction  is  caused  by  the  pressure 
of  the  clutch   facing  against  the 
FIG.  142. — Cone  clutch  used  on  metallic  balance  wheel.     The  pres- 

an  automobile.  gure    ^    duQ   ^   &  yery   gtrong    ^ 

spring.  The  spring  is  adjustable,  so  that  it  may  be  com- 
pressed in  case  of  slippage.  By  pushing  in  the  clutch  pedal, 
the  transmission  is  cut  off  from  the  source  of  power.  The 
cone  clutch  must  be  kept  free  from  dirt  and  should  be 


MECHANICAL  TRANSMISSION  OF  POWER  145 

given  an  occasional  application  of  Neat's  foot  oil  or  castor 
oil,  if  the  facing  is  leather.  Cone  clutches  must  be  re-faced 
eventually. 

The  cone  clutch  is  rapidly  being  superseded  by  the  multiple- 
disc  clutch  for  pleasure  cars.  The  multiple-disc  clutch  used 
on  the  Dodge  car  consists  of  seven  discs  held  in  engagement 
by  a  heavy  coil  spring.  The  driving  discs  (four  in  number) 
are  carried  by  six  studs  projecting  from  the  flywheel.  The 
driven  discs  (three  in  number)  are  carried  by  three  studs 
riveted  to  the  clutch  spider.  The  driving  discs  are  faced 
on  both  sides  with  wire- woven  asbestos  fabric.  The  driven 


FIG.  143. — Discs  from  a  Dodge  multiple-disc  clutch. 

discs  are  unfaced  steel.  If  the  clutch  pedal  is  pushed  in,  the 
spring  is  compressed  and  the  driving  discs  can  rotate  independ- 
ently of  the  driven  discs.  The  source  of  power  is  thus  cut  off 
from  the  transmission.  Modern  practice  favors  the  dry 
disc  clutch  as  it  gives  a  high  coefficient  of  friction.  Metal 
against  metal  gives  a  coefficient  of  .15  as  a  rule;  this  falls  to 
about  .07  if  the  surfaces  become  greasy.  Metal  against  fric- 
tion material  will  give  a  coefficient  of  .27  dry  and  .10  oiled. 
Thus  it  will  be  seen  that  metal  against  friction  material  tends 
to  produce  the  greater  friction.  The  disc  clutch  offers  a  large 
frictional  area  for  a  comparatively  small  clutch  diameter. 
Furthermore,  it  does  not  "grab."  Since  wear  is  even,  there 
is  less  danger  of  slippage  and  less  necessity  for  renewal  of 
facings. 

157.  Cams. — The  cam  is  an  inclined  plane  in  principle. 
The  shape  of  a  cam  will  depend  upon  its  use.     It  is  used  to 
10 


146 


MECHANICS 


operate  the  valves  of  an  internal  combustion  engine  and  to 
time  the  spark;  to  operate  punch  presses,  shears,  etc.  Cams 
require  good  lubrication,  as  they  work  with  a  great  deal  of 
friction.  They  are  necessary  in  certain  cases,  but  should  be 
avoided  whenever  possible.  The  cam  mechanism  for  lifting 
the  valves  of  a  Ford  motor  is  shown  in  Fig.  133.  The  six- 
sided  cam  in  Fig.  144  is  used  on  a  Chalmers  six  cylinder 
motor  for  timing  the  ignition.  Other  cars  use  a  similar 
arrangement  The  point  of  the  cam  strikes  the  contact  point 


CONTACT  POINT  L£V£« 


FIG.  144. — Automobile  spark  timer  operated  by  a  cam. 

lever,  interrupts  the  flow  of  current  in  the  primary  circuit  by 
separating  the  contact  points,  inducing  a  "jump"  spark  in  the 
secondary  circuit  (and  hence  in  the  spark  plug)  at  the  proper 
instant. 

158.  Links. — Power    is    often    transmitted    by    means    of 
link  work,  that  is,  by  means  of  levers,  cranks,  rods,  etc.     Links 
may  be  used  for  all  kinds  of  loads.     The  following  are  common 
examples  of  link  action:  the  foot-treadle  of  a  lathe;  the  con- 
necting rod  of  a  steam  engine;  the  valve  rod  of  a  Corliss  engine, 
etc.    Link  work  should  be  used  wherever  possible,  since  it 
possesses  the  least  friction  of  any  kind  of  transmission. 

159.  Screw    Threads. — The    projections    left    in    cutting 
helical  grooves  about  a  cylinder  are  called  threads.     External 
projections   form    male   threads;    internal   projections   form 


MECHANICAL  TRANSMISSION  OF  POWER  147 

female  threads.  Thus  a  bolt  is  said  to  have  male  threads  and 
a  nut  to  have  female  threads.  A  screw  is  single,  double  or 
triple  threaded  according  to  the  number  of  parallel  threads 
which  it  has.  The  lead  of  a  screw  is  the  linear  distance  that 
the  male  or  female  part  advances  during  one  complete  turn. 
The  lead  is  often  called  pitch  in  connection  with  single  threaded 
screws  and  is  equal  to  the  distance  between  the  centers  of  two 
adjacent  threads. 

Screw  threads  are  used  for  two  purposes :  binding  or  fasten- 
ing of  parts  and  transmission 
of    power.      Transmission 


screws  are  generally  square     '""'£'      '"^   '  y//         ™™ 
thread   or    acme    thread,    as       Fl(?  145.— Types  of  screw  threads: 

Such  threads  Operate  with  Square  thread  (a);  Acme  thread  (6); 
,  „  .  . .  , ,  . ,  i  •  i  American  or  Sellars  thread  (c) ;  English 

less  friction  than  other  kinds.    or  whitworth  thread  (d) . 
The  square  thread  screw  is 

used  on  the  jack  screw,  hand  presses,  etc.  The  acme  thread 
is  similar  in  construction,  but  the  sides  slope  slightly.  The 
acme  screw  is  stronger  and  is  used,  for  example,  on  the  lead 
screw  of  a  lathe.  It  is  especially  desirable  for  work  in  which 
the  screw  engages  with  a  split  nut.  The  split  nut  allows  the 
connection  to  be  broken  at  will.  Screws  of  the  "V"  type 
are  used  for  binding,  as  they  operate  with  a  large  amount  of 
friction. 

160.  Chains  and  Sprockets. — Chains  are  desirable  for 
transmitting  power  short  distances  when  slippage  is  to  be 
avoided.  They  are  extensively  used  for  bicycles,  automobile 
trucks,  electrically  driven  machine  tools,  etc.  Velocity 
ratios  of  1-8  and  linear  speeds  up  to  4,000  ft.  per  minute  are 
possible.  The  sprockets,  as  a  rule,  should  not  have  under  16 
teeth  and  the  distance  between  the  sprocket  centers  should 
not  be  less  than  1J/2  times  the  diameter  of  the  larger  sprocket 
or  more  than  12  ft.  As  all  chains  stretch,  provision  must  be 
made  for  shortening  the  chain  or  moving  the  sprockets  farther 
apart.  The  pressure  on  the  rivets  or  pins  of  a  chain  should 
never  exceed  700  Ib.  per  sq.  in.  Practical  tests  of  a  chain 
driven  bicycle  have  shown  an  efficiency  of  98  per  cent.  Block, 


148 


MECHANICS 


roller  and  Renold  (silent)  chains  are  the  ones  most  commonly 
used. 

The  Whitney  block  chain  is  shown  in  Fig.   146  and  the 
patent   connecting   link  for  the  same  is  shown  in  Fig.   147. 


FIG.  146. — Block  chain. 


FIG.  147. — Patent  connecting  link. 


FIG.  148. — Roller  chain  (detachable  type). 


FIG.   149. — Roller  chain  (riveted  type). 

Block  chains  should  not  exceed  800  ft.  per  minute  in  linear 
speed. 

The  Whitney  roller  chain  (detachable  type)  is  shown  in 
Fig.  148.  The  riveted  type  is  shown  in  Fig.  149.  It  should 
be  noted  that  the  detachable  type  provides  for  easy  removal 
or  separation  of  links.  Attention  is  also  directed  to  the  bush- 


MECHANICAL  TRANSMISSION  OF  POWER 


149 


ings  surrounding  the  rivets  or  pins.     This  type  of  a  chain  will 
operate  successfully  up  to  a  linear  speed  of  1,000  ft.  per  minute. 


FIG.     151.— Link-Belt    silent     chain 
showing  perfect  fit  to  sprocket. 


FIG. 


150. — Link-Belt  silent 
chain. 


The  silent  chain  (Fig.  150) 
is  manufactured  by  the  Link- 
Belt  Co.,  and  is  suitable  for 
light  or  heavy  duty.  It  is 
made  of  the  best  quality  steel 
and  consists  of  a  series  of 
leaves  or  plates  held  together 
by  case-hardened  bushings  and 
case-hardened  steel  pins. 
Linear  speeds  of  1,500  ft.  per 
minute  are  easily  possible  with 
this  chain.  Figure  151  shows 

how  perfectly  the  chain  fits  on  its   sprocket,    and   Fig.  152 
shows  a  silent  chain  in  actual  operation. 

161.  Pulleys   and  Belts. — Belts   are   extensively   used   for 
transmitting  power  up  to  60  ft.,  beyond  which  rope  transmis- 


FIG.  152.— Double  Link-Belt  silent 
chain  transmitting  325  hp. 


150 


MECHANICS 


f    '     \ 

ttr 


FIG.   153. — Open  drive  belt  (top  slack). 


Fio.   154. — Open  drive  belt  (bottom  slack). 


FIG.   155. — Open  drive  belt  with  fixed  idler. 

%^. 


T"  -•  •  1 

FIG.  156. — Open  drive  belt  with  swing  idler. 


FIG.  157. — Open  drive  belt  with  adjustable  idler. 


FIG.  158. — Crossed  drive  belt. 


MECHANICAL  TRANSMISSION  OF  POWER  151 

sion  is  cheaper.  They  are  suitable  for  low,  medium  or  heavy 
duty  work.  Leather  belts  are  best,  although  cotton,  hemp 
or  rubber  is  used  in  damp  places.  The  strongest  and  most 
durable  belts  are  made  of  oak-tanned  ox  hide.  Belt  tension 
should  not  exceed  350  Ib.  per  sq.  in.  of  cross  section.  Electrical 
transmission  has  replaced  the  belt  in  the  shop  to  a  certain 
extent;  yet  the  belt  will  always  retain  its  field  of  usefulness. 
All  faelts  tend  to  slip  or  creep,  resulting  in  wear  and  loss  of 
power.  By  using  top  slack  (Fig.  153)  a  greater  area  of  contact 
is  effected  between  pulley  and  belt  and  slippage  is  reduced. 


FIG.   159. — Compound  drive  belt. 


FIG.   160. — Tandem  drive  belt. 

Figure  154  illustrates  bottom  slack.  If  pulleys  are  not  lined 
up  exactly,  guides  are  necessary  to  keep  the  belt  from  coming 
off. 

Short  belts  generally  carry  a  fixed,  swing  or  adjustable  idler 
to  minimize  creep.  It  will  be  seen  from  Figs.  155,  156  and 
157  that  this  arrangement  provides  greater  frictional  area 
between  the  pulleys  and  belt. 

When  the  driven  pulley  must  rotate  opposite  in  direction  to 
the  driving  pulley,  crossed  drive  (Fig.  158)  is  used. 

Figures  159  and  160  show  the  compound  and  the  tandem 
drive  belt.  For  further  systems  of  belting  the  student  is 
referred  to  a  more  advanced  book. 


152  MECHANICS 

Below  is  a  photograph  of  a  30  in.  double  Ladew  "  Flint- 
stone"  belt  176  ft.  long.  This  belt,  according  to  the  manu- 
facturers, had  just  replaced  another  Ladew  belt  in  continuous 
use  for  18  years  and  crippled  through  accident,  not  wear. 


FIG.  161. — Ladew  30-in.  belt  (tandem  drive)  carrying  1,000  hp. 

Referring  to  Fig.  162,  A  is  a  driving  pulley  and  B  a  driven 
pulley,  direction  of  rotation  being  shown  by  arrows.  It  is 

evident  that  the  actual  driv- 
ing tension  is  largely  T2.    T\  is 
also  in   tension   and   tends   to 
retard  the    clockwise   rotation 
of  B.     Hence  the  total  effect- 
ive driving  tension  will  be  Tz  - 
Ti.     In  case  there  is  no  slip- 
page, the  horsepower  transmitted  is  as  follows: 
„      _  difference  in  tension  in  Ib.  X  velocity  of  belt  in  ft.  permin. 

33,000 

It  is  generally  assumed  that  a  single  belt  1  inch  wide  and 
running  1,000  ft.  per  minute  will  transmit  1  horsepower.  The 
rule  is  not  exact,  however.  The  following  table,  prepared  by 
the  Ladew  Belting  Co.,  will  be  found  conservative. 


MECHANICAL  TRANSMISSION  OF  POWER 


153 


HORSEPOWER  TRANSMITTED  BY  SINGLE  BELTS  AT  VARIOUS 

SPEEDS 


Belt  speed  ft.  per 
min. 

750 

1200 

1800 

2400 

3000 

3600 

4200  4800 

i 

5400 

6000 

Width,  inches 

Horsepower 

I 

1 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

2 

2 

4 

6 

7 

9 

11 

13 

15 

16 

18 

3 

3 

5 

8 

11 

14 

16 

19 

22 

25 

27 

4 

4 

7 

11 

15 

18 

22 

25 

29 

33 

36 

5 

5 

9 

14 

18 

23 

27 

32 

36 

41 

45 

6 

6 

11 

16 

22 

27 

33 

38 

44 

49 

55 

8 

7 

15 

22 

29 

36 

44 

51 

58 

65 

73 

9 

8 

16 

25 

33 

41 

49 

57 

65 

74 

82 

10 

9 

18 

27 

36 

45 

55 

64 

73 

82 

91 

12 

11 

22 

33 

44 

55 

65 

76 

87 

98 

109 

14 

13 

25 

38 

51 

64 

76 

89 

102 

115 

127 

16 

15 

29 

44 

58 

73 

87 

102 

116 

131 

145 

162.  Pulleys  and  Ropes. — Rope  transmission  is  used  exten- 
sively in  medium  and  heavy  duty  work,  for  both  main  and 
individual  drive,  when  the  distance  between  shafts  is  rela- 
tively great.  It  is  cheap  to  install  and  maintain,  noiseless 
and  very  flexible  in  application. 

Textile  ropes  are  generally  made  of  cotton,  hemp  or  manila 


FIG.  163. — Manila  transmission  rope. 

hemp.  Cotton  is  very  pliable  and  is  preferred  for  light  work. 
Manila  hemp  is  the  strongest  and  more  suitable  for  heavy 
work.  Textile  ropes  are  made  of  strands  (usually  four  and 
never  more  than  six)  wound  around  a  central  core.  A  special 
dressing  or  lubricant  is  used  to  overcome  internal  friction 
and  insure  long  life.  Figure  163  shows  a  four  strand  Dodge 


154 


MECHANICS 


"Firmus"  manila  transmission  rope.  A  four  strand  manila 
rope  1  inch  in  diameter  will  support  a  tension  of  nearly  9,000 
Ib.  If  the  diameter  is  doubled,  it  will  support  a  tension  of 
over  32,000  Ib. 

Wire  ropes  are  used  to  some  extent,  chiefly  for  hoisting 
purposes.     The  usual  form  consists  of  six  strands  with  'a  hemp 


FIG.  164. — Wire  transmission  rope. 


FIG.  165. — Rope  transmission  used  in  a  hydro-electric  plant. 

or  wire  core.  Individual  strands  are  composed  of  cast  steel 
wires  with  tensile  strengths  as  high  as  250,000  Ib.  per  sq.  in. 
There  are  generally  seven  or  nineteen  wires  to  a  strand.  By 
making  the  wires  less  in  diameter  and  increasing  their  number, 
greater  pliability  is  obtained.  Thus  a  nineteen-wire  strand  is 
more  pliable  than  a  seven-wire  strand  of  the  same  cross- 
section.  The  wire  rope  in  Fig.  164  has  six  strands,  each  strand 
being  made  up  of  nineteen  wires. 

Two  systems  of  rope  driving  are  in  use:  the  English  or 
multiple  system  and  the  American  or  continuous  system.     The 


MECHANICAL  TRANSMISSION  OF  POWER 


155 


English  system  provides  numerous  separate  ropes  connecting 
the  main  drive  pulley  to  the  various  driven  shafts,  or  separate 
ropes  connecting  the  main  drive  pulley  to  the  main  driven 
pulley,  the  latter  operating  the  distributing  shaft.  The 
American  system  employs  a  continuous  rope  connecting  the 
main  drive  pulley  to  the  driven  line  shaft,  to  which  shaft 
the  separate  units  are  attached.  In  case  of  a  broken  rope, 
the  English  system  will  not  be  out  of  commission  entirely;  a 
broken  rope  in  the  American  system  necessitates  an  immediate 
shut  down.  Figure  165  represents  a  rope  transmission 
installed  in  a  hydro-electric  plant.  The  total  power  trans- 
mission is  about  500  hp.  Attention  is  called  to  the  grooved 
pulleys.  The  grooves  act  as  guides  and  also  serve  to  increase 
the  friction  of  the  wheel  and  rope. 

163.  Friction  Wheels. — Friction  wheels  are  sometimes  used 


FIG.  166. — Friction  drive  formerly  used  on  a  Metz  automobile. 

for  light  and  medium  duty,  in  case  various  speed  changes 
are  desired.  There  are  three  types  of  friction  wheels:  the 
spur,  the  bevel  and  the  disc.  The  friction  surfaces  may  be 
of  metal,  wood,  fiber,  leather,  etc.  At  best,  friction  wheels  are 
unsatisfactory  and  have  lost  much  of  their  popularity.  It  is 
nearly  impossible  to  maintain  an  even  pressure  and  there  is 
great  stress  in  the  bearings.  Figure  166  represents  friction 
drive  applied  to  an  automobile.  A  steel  plate  is  connected 
directly  to  the  main  drive  and  engages,  at  right  angles,  with 


156  MECHANICS 

a  fiber  covered  wheel  connected  to  the  final  drive.  A  gearset 
is  thus  avoided,  changes  of  speed  and  direction  being  obtained 
by  sliding  the  driven  wheel  along  the  spline  shaft  on  which  it  is 
mounted.  Friction  transmission  has  attained  very  little 
prominence  in  the  automobile  world. 

164.  Toothed  Gears. — Toothed  gears  form  a  very  positive 
method  of  transmitting  power  from  one  shaft  to  another. 
Teeth  from  one  gear  engage  similar  teeth  in  another  gear, 

slippage  being  avoided.  For  low 
speeds  and  rough  work,  gears  may 
be  cast  from  iron.  For  high 


'*%  speeds    and    where    vibration    is 
eth'    undesirable,  gears  'are   cut  from 

steel.  Alloy  steel  is  generally  used  for  the  purpose.  In  order 
to  cut  down  chatter,  gears  are  sometimes  made  of  fiber  or 
rawhide.  Such  gears  will  not  stand  as  much  stress  as  steel 
gears  and  are  used  for  light  work,  such  as  the  timing  gears  of 
a  gasoline  motor.  Most  gears  employ  the  involute  tooth  (Fig. 
167),  other  shapes  having  become  nearly  obsolete.  The 
elements  of  gear  teeth  may  be  either  straight  lines  or  helical 
(curved)  lines. 

165.  Classification  of  Gears. — Gears  are  classified  with 
reference  to  the  elements  of  the  teeth  and  the  relative  position 
of  the  shafts.  Following  is  a  general  classification  of  gears  in 
common  use: 

1.  Spur  Gears. — Connect   parallel   shafts;   tooth   elements 
straight. 

2.  Bevel  Gears. — Connect  shafts  whose  axes  meet  if  pro- 
longed; tooth  elements  straight. 

3.  Worm  Gears. — Connect  shafts  that  are  not  parallel  and 
that  do  not  meet  if  prolonged;  consist  of  a  wheel  with  curved 
tooth  elements  in  engagement  with  a  worm  having  continuous 
screw-like  tooth  elements. 

4.  Helical  Gears. — Connect  parallel  shafts;  tooth  elements 
helical. 

5.  Spiral  Gears. — Connect  shafts  that  are  not  parallel  and 
that  do  not  meet  if  prolonged;  tooth  elements  helical. 


MECHANICAL  TRANSMISSION  OF  POWER 


157 


166.  Spur  Gears. — Spur  gears  have  teeth  whose  elements 
are  straight  lines.  They  are  used  to  connect  parallel  shafts 
that  are  close  together.  Spur  gears  are  extensively  used  for 
automobile  transmission,  lathes,  pumps,  hoisting  apparatus, 
etc. 

Figure  168  shows  the  ordinary  spur  gear  in  engagement  with 
its  pinion  (small  gear).  The  annular  or 'internal  gear  in  Fig. 
169  is  used  in  the  final  drive  of  certain  trucks.  Attention  is 


FIG.  168. — Spur  gear  and 
pinion. 


FIG.   169. — Annular  or  internal  gear  and 
pinion. 


called  to  the  fact  that  the  tooth  form  of  .the  internal  gear  is 
reverse  in  shape  to  that  of  the  pinion  gear. 

167.  Bevel  Gears. — Bevel  gears  have  teeth  whose  elements 
are  straight  lines  and  connect  shafts  whose  axes  meet  if 
sufficiently  prolonged.  The  axes  are  usually  at  right  angles, 
although  such  a  condition  is  not  necessary.  A  gear  bevelled 
at  an  angle  of  45°  is  called  a  mitre  gear.  If  the  teeth  of  a 
bevel  gear  are  curved,  it  is  called  a  curved-tooth  or  helical 
bevel  gear. 

The  bevel  gear  in  Fig.  170  has  straight  teeth.  The  helical 
bevel  gear  (Fig.  171)  is  commonly  used  in  the  final  drive  of 
most  pleasure  cars.  The  effect  of  the  curved  teeth  is  to 
obtain  a  smoother  and  quieter  drive  and  to  increase  the 
number  of  teeth  in  contact  at  a  given  instant. 


158  .  MECHANICS 

168.  Worm  Gears. — In  worm  gears  the  tooth  elements  are 
screw  like;  the  shafts  are  not  parallel -and  do  not  meet  if  pro- 
longed. Worm  gearing  affords  smooth  running  and  high 


• —  ^>«» 

FIG.  170. — Bevel  gear  and  pinion.  FIG.   171. — Helical  bevel  gear  and 

pinion. 


FIG.   172. — Worm  and  worm  gear. 

speed  reductions  and  has  been  used  with  great  success  on 
motor  trucks.  The  worm  and  worm  gear  shown  in  Fig.  172 
is  designed  for  rear  axle  drive,  but  can  be  used  for  other  pur- 
poses if  a  speed  reduction  is  desired.  The  particular  gear 
shown  here  is  phosphor-bronze  and  the  worm  is  hardened 


MECHANICAL  TRANSMISSION  OF  POWER 


159 


steel.  The  object  of  the  wide  tooth  angle  is  to  decrease 
friction.  This  makes  for  increased  efficiency  of  transmission 
and  longer  life  for  the  engaging  parts. 

169.  Helical  Gears. — The  tooth  elements  of  helical  gears 


FIG.  173 — Helical  gear  and  pinion. 


FIG.  174. — Double  helical 
or  herringbone  gear. 


are  curved  or  helical.  Helical  gears  connect  parallel  shafts. 
They  are  stronger  and  quieter  than  spur  gears  of  the  same 
width.  The  timing  gears  of  many  automobiles  are  now  of  the 
helical  type. 

Figure  173  shows  a  pair  of 
plain  helical  gears,  similar  to 
those  used  for  automobile  valve 
timing.  The  double  helical  or 
herringbone  gear  (Fig.  174)  is 
used  for  heavier  duty. 

170.  Spiral  Gears. — Spiral 
gears  have  curved  tooth  elements 
and  connect  shafts  that  are  not 
parallel  and  that  do  not  meet  if 
prolonged.  High  speed  reduc- 
tions are  possible  and  the  connection  of  shafts  in  different 
planes  is  also  brought  about.  Some  authorities  make  no 
distinction  between  helical  and  spiral  gears. 


FIG.   175. — Spiral  gears. 


160  MECHANICS 

Questions  and  Problems 

1.  State  nine  general  methods  for  the  mechanical  transmission  of 
power. 

2.  What  is  the  purpose  of  a  shaft?     Of  what  materials  are  shafts 
made? 

3.  What  is  meant  by  a  line  shaft?     Main  shaft?     Countershaft? 
Flexible  shaft?     Secondary  shaft?     Give  a  practical  example  of  each. 

4.  What  is  the  purpose  of  a  coupling?     State  the  difference  between 
a   fixed   and   flexible   coupling.     When   is   a   fixed   coupling -used?     A 
flexible  coupling? 

5.  What  is  a  universal  coupling?     For  what  purpose  is  it  used? 

6.  What  is  a  clutch?     For  what  kind  of  work  is  it  used?     Describe 
the  jaw,  cone  and  disc  clutch.     What  advantage  has  the  disc  clutch 
over  the  cone  clutch? 

7.  What  is  a  cam?     For  what  kind  of  work  are  cams  used ?     WThy  are 
they  avoided  whenever  possible? 

8.  What  is  a  link?     Give  various  examples  of  power  transmitted  by 
link  work.     Why  are  links  used  whenever  possible? 

9.  What  is  meant  by  a  thread?     What  is  a  male  thread?     Female 
thread?     What  is  the  difference  between  pitch  and  lead? 

10.  For  what  two  general  purposes  are  screw  threads  used?     Describe 
four  common  types  of  threads. 

11.  For  what  kind  of  work  is  chain  transmission  desirable?     Describe 
the  roller,  block  and  silent  chain  in  detail. 

12.  When  is  belt  transmission  used?     Why  is  "top  slack"  better  than 
"bottom  slack"?     Show,  using  diagrams,  various  belt  drives.     Describe 
three  kinds  of  idlers. 

13.  Given  the  difference  in  tension  and  the  belt  speed,  how  is  the  hp. 
transmitted  by  a  belt  determined? 

14.  Why  is  rope  transmission  much  used?     Describe  the  construction 
of  the  textile  and  wire  rope.     What  is  the  difference  between  the  Amer- 
ican and  English  system  of  rope  transmission? 

15.  Are  friction  wheels  satisfactory  in  power  transmission?     Explain. 

16.  What  is  a  gear?     How  are  gears  made?     What  is  meant  by  the 
tooth  element  of  a  gear?     Show  by  diagram  an  involute  gear. 

17.  Into  what  five  classes  may  gears  be  divided?     Define  each  care- 
fully and  point  out  the  kind  of  work  to  which  it  is  adapted. 


CHAPTER  XXI 
FLUIDS 

SECTION  I.     INTRODUCTORY 

171.  The  Three  Forms  of  Matter. — Matter  is  divided  into 
three  general  classes:  solids,  liquids  and  gases.  The  term 
fluid  is  used  to  include  both  liquids  and  gases.  Many  sub- 
stances are  capable  of  existing  in  any  of  the  three  forms. 
Water,  for  example,  may  readily  be  changed  to  ice  or  steam. 
Some  substances  are  hard  to  classify,  as  they  seem  to  occupy 
intermediate  positions.  In  other  words,  there  is  an  over- 
lapping which  makes  the  classification  only  approximate. 
Asphalt  is  ordinarily  a  solid,  but  on  a  very  hot  day  will  tend 
to  flow  and  is  then  considered  a  liquid.  The  exact  point 
at  which  matter  changes  from  a  solid  to  a  liquid,  liquid  to  a 
solid,  etc.,  is  often  difficult  to  determine. 

Solids  have  a  definite  shape  and  a  definite  volume  and  may 
be  compressed  to  a  certain  extent.  In  solids  the  cohesive 
force  between  the  molecules  is  relatively  great,  causing  the 
molecules  to  maintain  a  fixed  position  with  relation  to  each 
other.  The  latter  statement  in  no  way  contradicts  the  state- 
ment, made  earlier  in  the  book,  that  molecules  are  always  in 
vibration. 

Liquids  have  a  definite  volume  and  an  indefinite  shape 
and  are  practically  incompressible.  A  quart  of  water  will 
maintain  a  constant  volume,  but  will  vary  its  shape  to  meet  the 
shape  of  the  containing  vessel.  The  cohesive  force  between 
the  molecules  is  insufficient  to  cause  a  rigid  mass,  enabling 
the  molecules  to  move  about  with  more  or  less  freedom.  The 
incompressibility  of  liquids  is  made  use  of  in  transmitting 
pressure  through  them  from  one  point  to  another. 

Gases  have  neither  a  definite  shape  nor  a  definite  volume. 
There  seems  to  be  no  cohesive  force  between  the  molecules;  in 
11  161 


162  MECHANICS 

fact,  they  behave  as  if  they  are  in  mutual  repulsion.  A  gas 
like  ammonia,  if  introduced  into  a  room,  soon  expands  and 
penetrates  all  parts  of  the  room.  All  gases  are  readily 
compressible. 

172.  Density. — Density  is  the  mass  of  a  unit  volume  of  a 
substance.     In  the  f.p.s.  system,  density  is  measured  in  lb./ 
cu.  ft.     In  the  c.g.s.  system,  density  is  measured  in  g./cm.3 
For  example,  we  speak  of  the  density  of  water  as  62.5  lb./ 
cu.  ft.  or  1  g./cm.3     Density  is  determined  by  dividing  the 
weight  of  a  body  by  its  volume. 

n       .,          mass 

Density  =  — ; 

volume 

173.  Specific  Gravity. — Specific  gravity  is  a  number  express- 
ing how  many  times  a  substance  is  as  heavy  as  an  equal 
volume  some  other  substance  taken  as  a  standard.     For  solids 
and  liquids,  water  is  the  standard;  for  gases,  hydrogen  is 
usually  the  standard,  although  air  and  oxygen  are  sometimes 
used.     The  specific  gravity  of  sulphuric  acid  is  1.8.     This 
means  that  a  cu.  ft.  of  sulphuric  acid  weighs  1.8  times  as  much 
as  a  cubic  foot  of  water. 

weight  of  a  substance 
Specific  gravity  =       .  ,  — 7 —  7      7 —    —7 —  VT~~ 

weight  of  an  equal  volume  of  some  substance 

taken  as  a  standard 

174.  Pressure  Defined. — The  term  "pressure"  is  used  so 
commonly  in  physics  that  it  is  very  important  for  the  student 
to  have  a  clear  understanding  of  its  meaning. 

Pressure  =  force  acting  per  unit  area. 

Pressure  is  generally  given  in  lb.  per  sq.  in.  or  in  Kg.  per 
sq.  cm.  For  example,  the  manufacturers  of  the  Goodyear 
33  by  4  in.  cord  automobile  tire  recommend  that  the  tire  be 
kept  inflated  to  a  pressure  of  70  lb.  per  sq.  in.  or  5  Kg.  per  sq. 
cm.  The  same  pressure  may  be  expressed  more  briefly  as  70 
lb.  or  5  Kg.,  the  corresponding  units  of  area  being  understood. 

Questions  and  Problems 

1.  Discuss  the  three  forms  of  matter  in  detail. 

2.  Define  density.     Give  examples. 


FLUIDS  163 

3.  Define  specific  gravity.     Give  examples.     What  substances  are 
used  as  standards  in  specific  gravity? 

4.  Show  that  density  and  specific  gravity  are  numerically  the  same 
in  the  c.g,s.  system. 

5.  Explain  clearly  what  is  meant  by  pressure. 

6.  What  is  the  weight  of  a  piece  of  cast  gold  (sp.  gr.  =  19.3)  10  cm. 
on  a  side? 

7.  Two  hundred  cm3,  of  annealed  copper  weigh  1,778  g.     What  is 
its  density?     Specific  gravity? 

8.  A  sheet  of  silver  .001  cm.  thick  and  1  m.  square  weighs  105  g. 
What  is  the  density  of  silver? 

9.  A  vessel  is   just  filled  with  mercury   (sp.   gr.  =  13.6).     If  the 
mercury  weighs  20  lb.,  what  is  the  volume  of  the  vessel  in  cu.  ft.? 

10.  A  cast  iron  rod  (sp.  gr.  =  7.1)  is  1  in.  in  diameter  and  6  ft.  long. 
How  much  does  it  weigh? 

11.  A  cu.  ft.  of  cork  weighs  15  lb.     What  is  its  specific  gravity? 

12.  If  5  cu.  ft.  of  sulphuric  acid  weigh  562.5  lb.,  what  is  the  specific 
gravity  of  the  acid? 

13.  How  much  does  a  liter  of  sulphuric  acid  weigh?     Give  answer  in 
Kg. 

SECTION  II.  MECHANICS   OF  LIQUIDS 

175.  Pressure  in  Liquids  at  Rest. — It  is  a  matter  of  common 
experience  that  a  pressure  exists  in  all  liquids.    A  wooden  post 
thrust  down  into  a  body  of  water 
is  forced  back  so  that  a  portion  of  A 
the  post  is  out  of  water.    Divers  are 
unable  to  go  very  far  down  into  the 
water  due  to  the  pressure  to  which 
their  ears  are  subjected.   Submarines 
rarely  operate  more    than    200    ft.  Wafer 

below  the  surface,   on    account    of    Q 
the  tremendous  crushing  effect  on  FIG.  176. 

the  shell. 

Pressure  in  liquids  is  due  to  the  action  of  gravity  on  the 
various  particles  of  which  the  liquid  is  composed.  Assuming 
the  liquid  in  Fig.  176  to  be  at  rest,  it  is  evident  that  any 
selected  particle  is  in  equilibrium;  that  is,  the  forces  acting 
on  the  particle  are  balanced  and  their  resultant  is  zero.  If 
such  were  not  the  case,  the  particle  would  move  and  the  liquid 


164 


MECHANICS 


would  not  be  at  rest.     Hence,  in  a  liquid  at  rest,  pressure  at 
any  point  must  be  equal  in  all  directions. 

It  is  also  evident  that  the  pressure  exerted  by  the  liquid  in 
Fig.  176  is  normal  or  perpendicular  to  the  surface's  of  the 
container  ABCD.  Were  this  not  true,  a  component  of  the 
pressure  at  any  selected  point  would  cause  the  liquid  to  move 
along  the  surface,  contradicting  our  previous  assumption  that 
the  liquid  is  at  rest.  Liquid  pressure, 
then,  always  acts  at  right  angles  to  the 
retaining  surface. 

Liquid  pressure  on  a  horizontal  surface 
is  directly  proportional  to  the  depth  of  the 
surface  beneath  the  surface  of  the  liquid. 
Referring  to  Fig.  177,  suppose  we  have 
a  rectangular  vessel  1  X  1  X  10  in.  half 
filled  with  water.  The  pressure  on  the 
bottom  will  be  equal  to  the  weight  of 
the  water  or  1  X  1  X  5  X  .0361  =  .18 
Ib.  per  sq.  in.  If  the  vessel  is  entirely 
filled  with  water,  the  pressure  on  the 
bottom  will  be  1  X  1  X  10  X  .036  = 
.36  Ib.  per  sq.  in.  Hence  by  doubling 
the  depth,  we  have  doubled  the  pressure. 
Suppose,  further,  that  the  vessel  in 

Fig.  177  is  filled  with  mercury  (sp.  gr.  =  13.6).  The  pres- 
sure on  the  bottom  will  then  be  1  X  1  X  10  X  .036  X  13.6  = 
4.89  Ib.  per  sq.  in.,  or  13.6  times  as  much  as  if  the  vessel 
were  filled  with  water.  The  pressure  is  directly  proportional 
to  the  density  of  the  liquid. 

From  the  above  discussion,  we  may  summarize  as  follows: 
1..  At  any  selected  point,  the  pressure  exerted  by  a  liquid  is 
equal  in  all  directions. 

2.  Liquid  pressure  always  acts  at  right  angles  to  the  retaining 
surfaces. 

3.  The  pressure  in  each  layer  of  a  liquid  is  proportional  to 
the  depth. 

1 1  cu.  in.  water  weighs  .036  Ib. 


/ 

/ 

A 

5" 

J  . 

/ 

A 

s" 

Wa+er 

V 

/ 

k- 


FIG.   177. 


FLUIDS  165 

4.  With  different  liquids  and  the  same  depth,  the  pressure  is 
directly  proportional  to  the  density  of  the  liquid. 

Questions  and  Problems 

1.  Show  that  pressure  exists  in  liquids. 

2.  State  the  laws  of  liquid  pressure. 

3.  A  vessel  20  in.  high  is  filled  with  water.     Find  (a)  the  pressure  on 
the  bottom;  (6)  the  pressure  10  in.  down  from  the  surface. 

4.  Find    the    pressure  in  Problem  3  if  (a)   the  vessel  is  filled  with 
mercury;  (6)  with  sulphuric  acid. 

5.  If  the  vessel  in  Problem  3  is  filled  partly  with  mercury  and  partly 
with  sulphuric  acid,  find  the  pressure  on  the  bottom. 

6.  A  submarine  lies  so  that  a  portion  of  the  shell  is  200  ft.  below  the 
surface  of  the  ocean.     If  the  sp.  gr.  of  sea  water  is  1.03,  what  is  the 
pressure  on  the  shell? 

7.  A  depth  bomb  is  adjusted  to  explode  at  a  pressure  of  50  Ib.     At  what 
depth  will  the  explosion  take  place?     Solve  for  salt  water. 

8.  A  reservoir  is  70  ft.  deep.     Assuming  a  leakage  under  the  retaining 
wall  at  this  depth,  what  vertical  pressure  is  exerted  on  the  wall? 

9.  If  the  Titanic  sank  to  an  average  depth  of  3  miles,  to  what  pressure 
is  the  ship  subjected? 

176.  Transmission  of  Pressure  by  Liquids. — In  the  begin- 
ning of  the  chapter  it  was  stated  that  liquids  are  practically 
incompressible.  This  property  is  made 


L 


H20 


use  of  in  transmitting  pressure  from  one    *? 
point  to  another  through  liquids,  as  in 
the  hydraulic  press,  hydraulic  jack,  etc., 
in  which  a  tremendous  force  may  be 
exerted  with  only  a  small  applied  force. 

Suppose  we  have  a  vessel  (Fig.  178) 
containing  water,  fitted  with  a  plunger  FlG   178 

(P)  and  having  orifices  (a,  b  and  c)  of 

equal  cross-section.  If  the  plunger  is  pushed  down,  the  pres- 
sure of  the  plunger  will  be  transmitted  to  the  liquid  and  will 
cause  the  water  to  issue  from  the  orifices  in  streams  of  prac- 
tically equal  length.  This  shows  that  the  pressure  has  been 
transmitted  through  the  liquid  equally  in  all  directions. 

177.  Pascal's  Law.  —  The  following  law,  of  great  importance 
in  hydraulics,  was  enunciated  by  Pascal:1 

1  Blaise  Pascal  (1623-1662).    French  scientist,  mathematician,  philoso- 
pher and  author. 


166 


MECHANICS 


so* 

\  W    \ 

f* 

In. 

'fyffitf/fi, 

c 

ZOSq.In. 
C 

C/7 

Pressure  exerted  on  any  part  of  a  confined  liquid  is  trans- 
mitted equally  in  all  directions,  acts  with  equal  force  on  equal 
surfaces  and  at  right  angles  to  the  surfaces. 

Pascal's  Law  Illustrated.— C  and  c  (Fig.  179)  are  two 
communicating  cylinders  with  cross-sections  of  20  and  2  sq. 

in.  respectively.  P  and  p  are 
tight-fitting  pistons  (considered 
weightless).  The  space  between 
the  pistons  is  filled  with  some 
liquid  as  oil.  Suppose  a  force 
of  5  Ib.  is  acting  vertically  down 
on  p.  There  is,  then,  a  force  of 
FIG.  179. — Apparatus  for  iiius-  2.5  Ib.  per  sq.  in.  being  transmit- 

ted  to  the  liquid  by  the  plunger. 

Since  the  liquid  is  subject  throughout  to  a  pressure  of  2.5 
Ib.  per  sq.  in.,  a  pressure  of  2.5  Ib.  per  sq.  in.  is  transmitted 
to  every  sq.  in.  on  the  piston  P.  The  total  force  tending  to 
lift  P  is  therefore  2.5  X  20  or  50  Ib.  In  order  to  prevent 
P  from  moving  upward,  a  weight  of  50  Ib.  must  be  placed  on 
P.  In  the  above  case,  an  applied  force  of  5  Ib.  is  sufficient 
to  balance  a  load  of  50  Ib.  The  resisting  and  applied  forces 
are  directly  proportional  to  their  respective  cross-sections, 
Algebraically : 

W  :F  ::A:a 

178.  The  Hydraulic  Press. — Figure  180  is  a  simple  diagram 
showing  the  essential  parts  of  a  hydraulic  press.  The  press 
is  used  to  compress  cotton  into  bales  and  for  similar  work, 
whenever  it  is  desired  to  secure  a  large  force  by  the  application 
of  a  small  force.  The  gain  in  force  is  effected  by  making  P 
large  in  comparison  with  p  and  by  means  of  the  hand  lever 
operating  the  smaller  piston. 

The  total  crushing  effect  at  R  (friction  neglected)  is  found 
as  follows : 

R 


areaP      ac 

J-J       /\  /N         "  7 

area  p      ab 


Since  the  machine  is  considered  frictionless,  the  output 
at  R  will  equal  the  input  at  E.    Hence: 


FLUIDS 


167 


E  X  downward  distance  E  moves  —  R  X  upward  distance  P 

moves. 

The  student  should  note  that,  while  a  gain  in  force  is 


ter 
FIG.  180. — Hydraulic  press. 

secured,  there  is  no  gain  in  energy.     This  is  strictly  in  accor- 
dance with  the  law  of  conservation  of  energy  previously  studied. 
179.  Communicating   Columns. — It   is   a   common   saying 
that  "water  seeks,  its  own  level."     This  is  simply  another 


B 


Wafer 


Wafer 


1.8" 


FIG.  181. — Communicating 
columns. 


FIG.  182. — Heights  of  liquids 
are  inversely  proportional  to  their 
respective  densities. 


way  of  stating  that  water  contained  in  communicating  vessels 
will  come  to  rest  at  a  common  height.  Referring  to  Fig.  181, 
it  will  be  seen  that  column  A  contains  more  water  than  column 
B.  Nevertheless,  the  level  in  each  column  is  the  same.  This 
is  explained  by  Pascal's  Law.  Since  pressure  is  dependent 


168 


MECHANICS 


upon  depth,  it  is  clear  that,  in  order  for  the  pressure  at  the 
base  of  each  column  to  be  the  same  and  keep  the  liquid  in 
equilibrium,  the  liquid  in  the  separate  columns  must  have 
the  same  height. 

If  liquids  of  different  densities  are  used,  the  heights  will  be 
inversely  proportional  to  the  densities  of  the  liquids.  For 
instance,  Fig.  182  represents  a  column  of  water  and  a  column 
of  sulphuric  acid  separated  by  mercury  to  keep  them  from 
mixing.  The  mercury  is  first  put  in  and  will  come  to  rest 
with  both  surfaces  at  a  common  level.  Since  it  is  necessary 
.  that  the  level  of  the  mercury  shall  not  change,  the  pressure  of 
the  water  against  the  mercury  must  be  the  same  as  the  pres- 
sure of  the  acid  against  the  mercury.  As  sulphuric  acid  is 
1.8  times  as  dense  as  water,  this  condition  can  be  brought  about 
only  by  having  the  water  column  1.8  times  as  high  as  the  acid 
column.  Stated  algebraically: 

H  :h  ::d  :D, 

in  which  H  and  h  are  the  heights  corresponding  respectively  to 
the  densities  D  and  d. 

180.  Difference  Between  Pressure  and  Weight. — Pressure 
exerted  by  a  liquid  on  the  base  of  a  vessel  depends  in  no  way 

upon  the  shape  of  the  vessel: 
only  upon  the  depth  and  den- 
sity of  the  liquid.  Suppose 
(Fig.  183)  a,  b  and  c  to  be  three 
vessels  each  with  a  base  of  1 
sq.  in.,  10  in.  high  and  filled 
with  water.  The  pressure  at 
the  base  of  a  will  be  1  X  10  X 


10 


Wafer 


(a)  Cb)  (c) 

FIG.   183. — Diagrams  to  show  differ- 
ence between  pressure  and  weight. 


.036  or  .36  Ib.  per  sq.  in.  The 
weight  of  the  water  contained  in 
a  will  be  .36  Ib.  In  this  case  the  pressure  exerted  by  the  water 
and  the  weight  of  the  water  are  the  same.  Similarly,  we  find 
that  the  pressure  in  b  and  c  is  also  .36  Ib.  per  sq.  in.  Inspection, 
however,  shows  us  that  the  weight  of  the  water  in  b  is  less 
than  in  a;  also,  that  the  weight  of  the  water  in  c  is  greater  than 
in  a.  This  proves  that  pressure  and  weight,  are  different 


FLUIDS  169 

Questions  and  Problems 

1.  Why  is  a  liquid  better  than  a  gas  in  transmitting  pressure? 

2.  Give  an  experiment  to  show  that  liquids  transmit  pressure  equally 
in  all  directions. 

3.  State  and  illustrate  Pascal's  Law. 

4.  Describe  the  principle  of  the  hydraulic  press. 

5.  Explain  why  "water  seeks  its  own  level." 

6.  State  the  law  of  heights  for  communicating  columns  of  different 
densities. 

7.  Give  an  illustration  to  show  that  weight  and  pressure  are  not  the 
same. 

8.  Two  communicating  vessels   (A  and  B)  contain  water  and  are 
fitted  with  pistons.     The  piston  areas  are  10  and  1,000  sq.  in.  respec- 
tively.    If  a  downward  applied  force  of  50  Ib.  is  used  on  the  smaller 
piston,  find  (a)  the  pressure  transmitted  by  each  piston;  (6)  the  lifting 
effect  of  the  larger  piston. 

9.  Repeat  problem  8,  if  the  pistons  have  diameters  of  4  and  8  in. 
respectively. 

10.  The  small  piston  of  a  hydraulic  press,  as  shown  in  Fig.  180,  is 
2  in.  in  diameter  and  the  larger  piston  is  12  in.  in  diameter,     ac  is  36  in. 
and  ab  is  6  in.     Find  the  crushing  effect  at  R  if  E  is  40  Ib. 

11.  Find  R  in  problem  10,  if  a  mechanical  efficiency  of  75  per  cent, 
is  assumed. 

12.  Find  E  in  problem  10,  if  R  is  1,000  Ib.     Assume  an  efficiency  of 
70  per  cent. 

13.  In  Fig.  182,  if  the  water  is  5  in.  deep  and  the  cylinders  are  100  and 
10  sq.  in.  respectively  in  cross-section,  what  weight  of  sulphuric  acid 
must  be  used  to  keep  the  mercury  level  unchanged  ? 

181.  Total  Pressure  on  the  Various  Plane  Surfaces  of  a 
Vessel  Containing  a  Liquid. — Pressure  has  been  denned  as  the 
force  acting  per  sq.  unit  of  surface.  It  is  frequently  necessary 
to  compute  the  total  force  acting  against  an  area  exceeding  a 
sq.  unit  in  size.  This  total  force  is  called  total  pressure  and 
must  be  distinguished  carefully  from  pressure.  Total  pressure 
is  simply  the  sum  of  the  separate  pressures. 

(1)  HORIZONTAL  SURFACES 

We  have  seen  that  the  liquid  pressure  on  a  horizontal  sur- 
face is  equal  to  the  unit  area  X  the  height  of  the  liquid  X 
the  density  of  the  liquid.  It  follows,  therefore,  that: 


170 


MECHANICS 


The  total  liquid  pressure  acting  upon  a  horizontal  surface 
equals  the  area  of  the  surface  X  the  height  of  the  liquid  above 
the  surface  X  the  density  of  the  liquid. 

(2)   VERTICAL  SURFACES 

In  dealing  with  vertical  surfaces,  we  must  take  into  account 
that  the  pressure  against  the  surface  is  not  constant,  but 
increases  with  the  depth.  The  total  pressure  will  then  be  the 
area  of  the  surface  X  the  average  pressure  of  the  liquid.  We 
may  state  the  rule  for  total  pressure  as  follows: 

The  total  liquid  pressure  acting  upon  a  vertical  surface 
equals  the  area  of  the  surface  in  the  liquid  X  the  vertical  dis- 
tance from  the  center  of  gravity  of  the  surface  in  the  liquid  to 
the  top  of  the  liquid  X  the  density  of  the  liquid. 

(3)  SURFACES  NEITHER  HORIZONTAL  NOR  VERTICAL 

Since  liquid  pressure  always  acts  at  right  angles  to  any 
surface,  the  rule  stated  in  (2)  is  applicable  to  surfaces  which 
are  not  vertical  or  horizontal. 

The  student  should  note  that  the  law  in  (2)  holds  in  (1) 
as  well.  Hence  the  rule  given  in  (2)  for  vertical  surfaces 
may  be  used  for  any  plane  surface  in  the  solution  of  problems. 

182.  Center  of  Pressure. — Since  pressure  against  a  sub- 
merged surface  always  acts  at  right  angles  to  the  surface, 


FIG.  184. — Center  of  pressure  for  a  horizontal  surface. 

we  may  conceive  of  the  total  pressure  as  being  made  up  of 
countless  parallel  forces.  The  total  pressure  is  the  resultant  of 
the  separate  forces. 


FLUIDS 


171 


I  * 


Wafer 


-1 


C.R 


•3f 


Figure  184  (a)  represents  a  horizontal  surface  acted  upon 
by  parallel  forces.  Since  the  forces  are  equal,  we  may  assign 
a  value  of  "/"  to  each.  As  there  are  25  forces,  the  resultant 
is  evidently  25/.  If  we  replace  the  separate  forces  by  a  single 
force  of  25/  applied  at  the  center  of  area  of  the  surface  as 
shown  in  Fig.  184(6),  the  effect  will  be  the  same  as  in  Fig. 
184 (a).  The  point  at  which  the  separate  parallel  liquid  forces 
acting  against  a  submerged  surface  may  be  replaced  by  a  single, 
resultant  force,  is  called  the  center  of  pressure  for  the  surface. 
For  horizontal  surfaces,  the  center  of  pressure  is  always  at 
the  center  of  gravity  of  the  surface. 

In  case  the  submerged  surface  is  vertical  and  rectangular, 
the  center  of  pressure  will  be  lo- 
cated below  the  center  of  gravity. 
This  is  due  to  the  fact  that  the 
pressure  increases  with  the  depth 
as  shown  in  Fig.  185.  Referring 
to  Fig.  185,  we  may  represent  the 
separate  forces  by/,  2/,  3/  and  4/. 
The  resultant  force  is  10  /.  It  will 
be  seen,  by  inspection,  that  the 
resultant  force  (10f)  must  be 
applied  below  the  center  of  gravity  to  produce  the  same  effect 
as  the  separate  forces.  In  this  case,  the  center  of  pressure  is  % 
of  the  vertical  distance  from  the  top  of  the  surface  to  the  base. 

For  surfaces  not  horizontal,  and  submerged  so  that  the  top 
is  in  line  with  the  level  of  the  liquid,  the  center  of  pressure  is 
located  as  follows: 

1.  Rectangular  Surfaces. — Center  of  pressure  is  %  the  vertical  distance 
from  the  top  to  the  base. 

2.  Triangular  Surfaces  with  Base  Horizontal  and  Vertex  at  Top. — Center 
of  pressure  is  %  the  distance  from  the  vertex  to  the  mid-point  of  the 
base. 

3.  Triangular  Surfaces  with  Base  at  Top. — Center  of  pressure  is  %  the 
distance  from  the  mid-point  of  the  base  to  the  vertex. 

183.  Dams  and  Retaining  Walls. — Great  care  must  be 
exercised  in  the  construction  of  dams  and  retaining  walls. 


FIG.  185. — Center  of  pressure  for 
a  rectangular  vertical  surface. 


172 


MECHANICS 


The  most  important  feature  of  a  dam  or  retaining  wall  is 
the  foundation.  Poorly  constructed  foundations  allow  water 
to  leak  in,  impairing  the  base  and  at  the  same  time  exerting 
an  upward  pressure.  This  may  result  in  a  failure  of  the 
structure. 

Assuming  a  proper  foundation,  a  dam  or  retaining  wall 
is  liable  to  failure  in  three  ways:  (1)  sliding  horizontally; 
(2)  crushing  of  materials;  (3)  overturning. 

1.  Stability   against   sliding   is   obtained    by   making   the 


FIG.  186.— Dam  A-B-C-D  will 
not  overturn  around  A  if  moment 
W  X  di  exceeds  moment  P  X  di. 


FIG.   187. — Transverse  outline  of  the 
Ashokan  dam. 


structure  sufficiently  heavy  and  allowing  plenty  of  friction 
at  the  base. 

2.  Stability  against  crushing  is  obtained  by  keeping  the 
compression,  to  which  the  material  is  subjected,  less  than  its 
crushing  strength. 

3.  Stability  against  rotation  is  obtained  by  making  the 
structure  heavy  and  thick  at  the  base,  causing  the  stabilizing 
moment  to  exceed  the  turning  moment.     For  example,  in 
Fig.  186,  the  dam  A  BCD,  if  overturned,  would  swing  about  A 
as  a  fulcrum.     The  moment  tending  to  overturn  the  dam  is 


FLUIDS  173 

P  X  d2.  P  is  the  resultant  liquid  pressure  applied  at  the 
center  of  pressure  and  d2  is  the  moment  arm  or  perpendicular 
distance  to  the  fulcrum  A.  The  stabilizing  moment  is  W 
X  di.  W  is  the  weight  of  the  dam  concentrated  at  the  center 
of  gravity  and  di  is  the  corresponding  moment  arm.  If  W 
X  dL  exceeds  P  X  d2,  there  will  be  no  rotation  about  A. 

Figure  187  shows  the  outline  of  a  transverse  section  of  the 
Ashokan  dam,  a  part  of  the  water  system  for  Greater  New 
York.  The  masonry  part  is  built  of  irregular  rocks  cemented 
with  poured  concrete  and  faced  with  concrete  blocks.  It  is 
1,000  ft.  long  and  220  ft.  high.  The  width  of  base  is  190.2 
ft.  and  the  width  at  the  top  is  26.3  ft. 

Questions  and  Problems 

1.  Distinguish  between  pressure  and  total  pressure. 

2.  State  the  rule  for  finding  the  total  liquid  pressure  acting  against  a 
plane  surface. 

3.  What  is  meant  by  center  of  pressure?     Where  is  the  center  of 
pressure  located  in  a  submerged  horizontal  surface?     Submerged  vertical 
surface? 

4.  Why  must  the  base  of  a  dam  or  retaining  wall  be  constructed 
carefully  ? 

5.  Assuming  a  proper  foundation,  mention  three  causes  why  a  dam 
or  retaining  wall  is  liable  to  fail. 

6.  Explain  the  precautions  taken  to  guard  against  failure  as  indicated 
by  the  answer  to  Problem  5. 

7.  Give  a  brief  description  of  the  Ashokan  Dam. 

8.  A  swimming  tank  is  50  ft.  long,  35  ft.  wide  and  8  ft.  deep.     If  the 
tank  is  filled  with  water,  find  (a)  the  weight  of  the  water;  (6)  the  total 
pressure  on  the  bottom;  (c)  the  total  pressure  on  each  side;  (d)  the  total 
pressure  on  each  end. 

9.  A  dam  is  100  ft.  long  and  holds  back  a  body  of  water  2  miles  in 
length.     If  the  water  is  30  ft.  deep,  find  the  total  pressure  that  the  dam 
must  withstand. 

10.  A  gate  in  a  dam  is  1  m.  wide  and  2  m.  high.     The  top  of  the  gate  is 
4  m.  below  the  water  level.     Find  the  total  pressure  against  the  gate  in 
Kg. 

11.  A  swimming  tank  40  ft.  long  and  15  ft.  wide  is  entirely  filled  with 
water.     The  water  is  4  ft.  deep  at  one  end  and  slopes  evenly  to  a  depth  of 
6  ft.  at  the  other  end.     Find  (a)  the  total  pressure  on  the  bottom;   (6) 
the  total  pressure  on  each  end. 


174 


MECHANICS 


12.  A  submarine  lies  so  that  a  certain  horizontal  section  2  m.  square  is 
50  m.  below  the  level  of  the  water.     Find  the  total  pressure  against  the 
section  in  Kg.     Density  of  sea  water  is  1.03. 

13.  A  pressure  gauge  applied  to  a  city  main  registers  75  Ib.     How 
high  above  the  gauge  is  the  water  in  the  distributing  reservoir? 

14.  A  covered  rectangular  tank  is  6  ft.  long,  4  ft.  wide  and  4  ft.  deep. 
A  tube  6  X  4  in.  and  24  in.  high  extends  up  from  the  cover.     If  the  tank 
and  tube  are  filled  with  water,  find  (a)  the  total  weight  of  the  water; 
(6)  the  total  pressure  on  the  bottom  of  the  tank;  (c)  the  total  pressure  on 
each  end  of  the  tank;  (d)  the  total  pressure  on  each  side  of  the  tank; 
(e)  the  total  pressure  on  each  surface  of  the  tube. 

184.  Waterwheels. — Waterpower  has  been  used  for  cen- 
turies.    If  available,  it  is  both  convenient  and  economical. 


FIG.  188. — Overshot  water 
wheel. 


FIG.   189. — Undershot  water  wheel. 


Many  cities  owe  their  existence  and  growth  to  natural 
waterfall.  Water  furnishes  the  motive  power  for  mills,  fac- 
tories, electric  plants,  etc.  Various  types  of  waterwheels  are 
in  use.  Modern  practice  favors  the  Pelton  and  the  turbine, 
although  the  old-fashioned  overshot  and  undershot  wheels  are 
still  used  to  some  extent. 

The  overshot  wheel  (Fig.  188)  depends  chiefly  on  the  potential 
energy  of  the  water  flowing  into  the  buckets;  but  some  energy 
is  derived  from  the  impact  of  the  water  against  the  buckets. 
It  has  been  used  mostly  in  mountainous  places  where  the 
fall  is  good,  but  the  actual  volume  of  water  small.  The  effi- 
ciency of  the  overshot  wheel  is  often  as  high  as  90  per  cent. 
The  loss  is  due  to  the  friction  of  the  bearings,  etc.,  as  well  as  to 


FLUIDS 


175 


the  water  which  misses  the  buckets  or  which  is  spilled  from 

them. 

The  undershot  wheel  (Fig.  189)  is  used  in  flat  regions  where 

there  is  a  large  volume  of  water  with  poor  fall.     It  depends 

entirely  upon  the  kinetic  energy  of  the 

water.     Not  over  30  per  cent,  efficiency 

may  be  expected  from  a  wheel  of  this 

type. 

The  Pelton  wheel  (Fig.  190)  has  met 

with  much  success.     It  is   ordinarily 

used  in  water  motors  for   household 

purposes  or  for  other  purposes  where 

a   small  horsepower  is  desired.     Pelton  wheels   have  been 

built,  however,  capable  of  delivering  several  thousand  hp. 

Water  at  high  pressure  is  directed  against  the  cup-shaped 

buckets  and  a  large  part  of  its  kinetic  energy  is  transferred 

to  the  wheel.  The  efficiency  of 
the  Pelton  will  often  run  over  80 
per  cent. 

The  prevailing  type  of  water- 
wheel  today  is  the  turbine.  The 
turbines  at  Niagara  develop  about 


FIG.   190. — Pelton  water 
wheel. 


.-Generator 


Water 


Turbine 


Tail  Water 
FIG.  191. — Water  turbine. 


Wafer 


FIG.  192. — Diagram  show- 
ing fixed  and  movable  blades 
of  a  water  turbine. 


5,000  hp.  each.  Figure  191  shows  a  typical  installation.  Water 
enters  the  penstock  and  flows  down  into  the  turbine  case.  The 
turbine  wheel  rotates  horizontally  in  an  inner  case.  The  fixed 


176  MECHANICS 

guides  (Fig.  192)  direct  the  water  .against  the  movable  blades 
at  the  proper  angle,  rotating  the  wheel  which  is  connected  to 
the  electric  generator  by  a  vertical  shaft.  After  imparting  a 
large  part  of  its  energy  to  the  turbine,  the  water  drops  down 
into  the  tail  race.  The  amount  of  water  flowing  through  the 
case  is  controlled  by  means  of  a  gate  valve.  The  maximum 
efficiency  will  be  about  90  per  cent.  The  energy  expended 
upon  a  turbine  per  second  is  equal  to  the  product  of  the 
weight  of  the  water  flowing  through  it  per  second  and  the 
height  of  the  water  above  the  bottom  of  the  pit. 

Questions  and  Problems 

1.  Discuss  waterpower  as  a  source  of  energy. 

2.  Give  the  construction  of  an  overshot  waterwheel.     Explain  how  it 
gets  its  energy.     What  is  the  maximum  efficiency  possible  with  a  wheel 
of  this  type? 

3.  Repeat  as  above  for  an  undershot  waterwheel. 

4.  Describe  the  construction  and  operation  of  a  Pelton  waterwheel. 
For  what  kind  of  work  is  it  generally  used?     What  is  the  maximum 
efficiency  to  be  expected? 

6.  Describe  the  construction  and  operation  of  a  modern  turbine  power 
plant.  Why  is  the  turbine  so  efficient?  How  is  the  energy  per  second 
expended  upon  a  turbine  determined? 

6.  The  hp.  of  a  Niagara  turbine  is  5,000.     The  pit  is  136  ft.  deep. 
How  much  water  passes  through  the  turbine  per  minute,  if  the  efficiency 
is  85  per  cent.  ? 

7.  Repeat  Problem  6,  using  efficiencies  of  (a)  80  per  cent. ;  (6)  90  per 
cent. 

185.  Buoyant  Force  of  Liquids. — It  is  common  knowledge 
that  bodies  weigh  less  in  water  than  in  air. 
A  heavy  stone  may  be  moved  much  more 
easily  in  water  than  in  air.  Floating  bodies 
lose  their  entire  weight  when  placed  in  a 
liquid.  The  following  paragraph  will 

Wafer   D^bT>('         explain   the  buoyant    or   lifting   effect  of 

FIG.  193.— Diagram   liquids. 

of  Referring  to  Fig.  193 (a),  we  have  a  rect- 
angular solid  A  BCD  (side  view  shown) 
immersed  in  some  liquid  as  water.  The  total  pressure  push- 
ing the  body  downward  is  equal  to  the  weight  of  a  column 


FLUIDS  177 

of  the  liquid  the  size  of  the  volume  FADE.  The  total  upward 
pressure  at  the  base  of  the  body  is  equal  to  the  weight  of 
a  column  of  water  the  size  of  the  volume  FBCE.  The  buoy- 
ant force  of  the  liquid  is  equal  to  the  difference  between 
the  total  upward  and  downward  pressures  or  the  weight  of 
a  column  of  liquid  the  size  of  ABCD.  Since  FBCE  =  FADE 
+  ABCD,  algebraically: 

FADE  +  ABCD  -  FADE  =  ABCD 

From  the  above,  we  see  that  the  loss  of  weight  of  the  body 
is  equal  to  the  weight  of  the  liquid  displaced.  By  similar 
reasoning,  since  liquids  are  practically  the  same  density 
throughout,  it  may  be  proven  in  Fig.  193(6)  that  the  buoyant 
force  will  remain  the  same  if  the  body  is  farther  down  in  the 
liquid.  Hence  we  may  consider  buoyancy  independent  of 
depth. 

The  relation  between  the  loss  of  weight  of  a  body  and  the 
liquid  displaced  was  discovered  by  Archimedes1  and  is  called 
Archimedes'  Principle. 

Archimedes1  Principle. — The  loss  of  weight  of  a  body  immersed 
in  a  liquid  is  equal  to  the  weight  of  the  liquid  displaced. 

Since  a  floating  body  loses  all  of  its  weight  when  placed  in  a 
liquid,  the  weight  of  the  liquid  displaced  must  be  equal  to  the 
original  weight  of  the  body. 

Archimedes'  principle  furnishes  a  very  convenient  method  of 
determining  the  specific  gravity  of  solids.  For  bodies  heavier 
than  water,  the  procedure  is  very  simple.  The  body  is 
weighed  in  air  and  then  in  water.  The  specific  gravity  is  equal 
to  the  weight  of  the  body  in  air  divided  by  the  loss  of  weight  in 
water.  In  case  of  floating  bodies,  it  is  necessary  to  attach 
a  sinker  to  the  body  of  sufficient  size  to  submerge  it.  The 
following  problems  will  illustrate  each  method. 

1  Archimedes  (287-212  B.C.).     Greatest  mathematician  of  antiquity. 
Born  in  Syracuse,  Sicily.     First  determined  the  value  of  TT  and  the  area 
of  a  circle.     Discovered  the  laws  of  notation  and  of  the  lever.     Killed 
by  a  Roman  soldier  during  the  sack  of  Syracuse. 
12 


178 


MECHANICS 


1.  A  piece  of  wrought  iron  weighs  10  Ib.  in  air  and  8.73  Ib.  in  water. 
What  is  the  specific  gravity  of  the  iron?  It  is  evident  that  the  loss  of 
weight  of  the  iron  in  water  is  equal  to  the  weight  of  the  water  displaced. 
Therefore: 

10 


Sp.  gr.  wrought  iron  = 


10  -  8.73 


7.87 


2.  A  block  of  paraffine  weighs  2  Ib.  in  air.  It  is  attached  to  a  sinker 
of  sufficient  size  to  submerge  it.  First  the  block  and  sinker  are  weighed 
with  the  sinker  submerged  and  the  paraffine  in  the  air.  This  weight  is 
5  Ib.  Next  the  weight  is  taken  with  both  submerged.  This  is  found 
to  be  2.76  Ib.  Hence  the  buoyant  force  of  the  water  on  the  paraffine  or 
the  weight  of  a  volume  of  water  the  size  of  the  paraffine  is  5  -  2 . 76  or 
2.24  Ib.  Therefore: 


Sp.  gr.  paraffine  =  7 

o  — 


.89 


186.  The 


1000 


?000 


FIG.  194.— Con- 
slant-weight  hy- 
drometer. 


Hydrometer. — A  floating  body  displaces  its 
own  weight  of  a  liquid.  If  the  floating  body 
is  placed  in  liquids  of  different  densities,  the 
volumes  displaced  in  the  several  cases  will  be 
inversely  proportional  to  the  respective  den- 
sities. The  hydrometer,  an  instrument  for 
measuring  the  specific  gravity  of  liquids,  is 
constructed  to  work  on  this  principle.  The 
hydrometer  shown  in  Fig.  194  consists  of  a 
glass  tube  with  two  bulbs.  The  stem  and 
upper  bulb  contain  air;  the  lower  bulb  is 
loaded  with  shot  so  that  the  apparatus  will 
float  upright.  The  instrument  shown  here 
is  for  liquids  heavier  than  water.  When  placed 
in  water,  the  graduation  mark  1,000  is  even 
with  the  water  level.  Since  the  specific  grav- 
ity of  water  is  1,  it  is  clear  that  we  must 
read  the  1,000  graduation  mark  as  1.  When 
placed  in  sulphuric  acid,  the  graduation  mark 
corresponding  to  the  acid  level  will  be  1,800, 
showing  that  the  specific  gravity  of  sulphuric 
acid  is  1.8.  The  lower  graduation  of  the 
hydrometer  here  described  is  2,000,  giving  a 
1  to  2.  The  hydrometer  used  for  liquids 

1,000  at  the 


range   of  from 

lighter  than  water  is  usually  graduated  from 


FLUIDS  179 

bottom  to  600  at  the  top.  When  working  exclusively  with 
certain  liquids,  special  hydrometers  are  provided  which  are 
more  sensitive.  The  hydrometer  used  to  test  milk  is  called 
a  lactometer;  for  alcohol,  an  alcoholmeter,  etc. 

Questions  and  Problems 

1.  What  is  meant  by  the  "buoyant  force  of  liquids?" 

2.  State  Archimedes'  principle.     Does  it  apply  to  floating  bodies? 

3.  An  aluminum  block  is  4  ft.  long,  4  ft.  wide  and  3  ft.  high.     If  the 
block  is  submerged  in  water,  prove  that  the  total  upward  pressure  acting 
on  the  block  minus  the  total  downward  pressure  on  the  block  is  equal  to 
the  weight  of  the  water  displaced. 

4.  A  piece  of  wood  4  in.  square  floats  with  %  of  its  volume  submerged. 
Find  (a)  the  weight  of  the  wood;  (6)  the  specific  gravity  of  the  wood. 

6.  A  loaded  auto  truck  is  driven  aboard  a  small  ferry.  The  ferry  is 
30  ft.  long  and  15  ft.  wide.  If  the  ferry  sinks  IK  in.,  what  is  the  weight 
of  the  truck? 

6.  A  body  weighs  20  Ib.  in  air  and  10  Ib.  in  water.     Find  its  specific 
gravity. 

7.  A  hollow  ball  of  cast  iron  weighs  500  g.     What  must  be  its  volume 
in  cm3,  in  order  for  it  to  remain  suspended  if  submerged  in  water? 

8.  What  must  be  the  volume  of  the  ball  in  problem  7,  if  kerosene 
(sp.  gr.  =  .79)  is  used  instead  of  water? 

9.  A  sinker  weighs  2,000  g.  in  air,  1,600  g.  in  water  and  1,588  g.  in 
salt  water.     Find  (a)  the  specific  gravity  of  the  sinker;  (6)  the  specific 
gravity  of  the  salt  water. 

10.  A  piece  of  paraffine  (sp.  gr.  =  .89)  is  1  ft.  square.     How  far  will 
it  sink  in  a  salt  solution  whose  specific  gravity  is  1.03? 

11.  A  wooden  hydrometer  1  X  1  X  30  cm.  sinks  27  cm.  in  water  and 
25  cm.  in  a  solution  of  copper  sulphate.     Find  the  specific  gravity  of  the 
solution. 

12.  A  body  weighs  2  Kg.  in  air.     It  is  attached  to  a  sinker  which  alone 
weighs  3,000  g.  in  water.     The  sinker  is  just  sufficient  to  submerge  body. 
Find  the  specific  gravity  of  the  body. 

13.  A  piece  of  wood  weighs  2  Kg.  in  air.     It  is  fastened  to  a  sinker  and 
the  sinker  and  wood  when  suspended  together  under  water  weigh  3.5 
kg.     The  sinker  alone  in  water  weighs  6  Kg.     Find  the  specific  gravity 
of  the  wood. 

14.  A  uniform  glass  tube  5  ft.  long  has  an  outside  cross-section  of 
4  sq.  in.     It  is  loaded  with  mercury  at  the  closed  end.     The  tube  and 
mercury  together  weigh  15  Ib.     How  many  cu.  in.  of  cork  (sp.  gr.  = 
.25)  must  be  fastened  to  the  tube  so  that  it  will  float  upright  with  the 
top  1  ft.  out  of  water?     The  cork  must  be  entirely  submerged. 


180  MECHANICS 

SECTION  III.    MECHANICS  OF  GASES 

187.  Characteristics  of  Gases. — We  have  seen  in  Sect.  I 
that  gases  have  neither  a  definite  shape  nor  a  definite  volume ; 
that  they  readily  admit  of  compression ;  and  that  they  tend  to 
expand  indefinitely  if  unconfined.     We  may  now  state,  in 
addition,  that  gases  follow  the  laws  of  Pascal  and  Archimedes 
and  that  the  pressure  exerted  by  a  gas  at  any  point  is  equal 
in  all  directions.     Gases  expand  when  heated  and  contract 
when  cooled.     If  a  confined  gas  is  heated,  it  will  undergo  an 
increase  in  pressure.     If  external  pressure  is  applied  to  a 
confined  gas  (e.g.,  the  mixture  of  gasoline  vapor  and  air  in  a 
gasoline  motor  cylinder),  the  volume  of  the  gas  will  become 
less;  if  the  pressure  is  removed,  the  gas  will  expand  to  its 
former  volume.     The  above  changes  will  be  studied  in  detail 
later  in  the  chapter. 

188.  The  Atmosphere. — The  earth  is  surrounded  by  a  vast 
sea  of  atmosphere.    Like  liquids,  the  atmosphere  has  weight 
and  exerts  pressure.     At  sea  level  the  air  exerts  a  pressure  of 
14.7   Ib./sq.   in.  or   1033.6  g./cm2.     For  approximate  work 
we  may  take  the  pressure  as   15  Ib./sq.  in.  or  1  Kg./cm2. 
It  is  supposed,  since  the  pressure  decreases  with  altitude,  that 
the  extent  of  the  atmosphere  is  limited.     It  is  believed  that 
no  atmosphere  exists  at  a  distance  of  about  500  miles  above 
the  earth.     Air  is  a  mixture,  containing  about  20  per  cent, 
by  volume  of  oxygen  and  75  per  cent,  by  volume  of  nitrogen, 
as  well  as  small  quantities  of  carbon  dioxide,  water  vapor, 
etc.     At  average  sea  level  at  a  temperature  of  0°C.,  the  density 
of  air  is  .0807  Ib./cu.  ft.  or  .001293  g./cm3.     Thus  12  cu.  ft. 
of  air  weigh  approximately  a  pound.     A  room  may  easily 
contain  a  ton  of  air.     If  the  atmosphere  had  the  same  density 
at  all  altitudes  as  at  sea  level,  it  would  extend  upward  about 
5  miles.     This  height  is  called  the  height  of  the  homogeneous 
atmosphere. 

189.  Proof  that  Air  has  Weight. — We  may  prove  that  air 
has  weight  by  the  following  simple  experiment.     Figure  195 
represents  a  two  liter  flask  attached  to  the  bell  glass  of  an  air 


FLUIDS 


181 


pump  as  shown.  First  the  flask  and  the  air  therein  contained 
are  weighed.  The  air  is  then  exhausted  and  the  container 
minus  the  air  is  weighed.  The  difference  in  the  weights 
recorded  is  the  weight  of  the  two  liters  of  air  formerly  con- 
tained in  the  flask.  If  the  air  pump  is  efficient,  the  true 
weight  of  the  air  will  be  found  very  closely  and  will  be  approxi- 
mately 2.6  g. 

190.  Proof  that  Air  Exerts  Pressure. 
It  is  a  matter  of  every-day  knowledge 
that  the  air  exerts  pressure.  Auto- 
mobile tires,  foot  balls,  etc.,  depend 
upon  air  pressure.  In  constructing  the 
Hudson  Tubes  under  the  North  River, 


FIG.  195. — Appa- 
ratus to  show  that 
air  has  weight. 


FIG.  196, — Appa- 
ratus to  show  that 
air  exerts  pressure. 


compressed  air  was  used  to  prevent  the  water  from  entering  the 
tubes  while  work  was  under  way.  The  following  demonstra- 
tion proves  most  strikingly  that  air  exerts  pressure.  A  wood 
funnel  is  fitted  tightly  into  a  bladder  glass  (Fig.  196)  and  the 
bladder  glass  is  attached  in  turn  to  the  bell  glass  of  an  air 
pump.  Mercury  is  poured  into  the  funnel  and  the  air  is 
withdrawn  from  beneath  it.  The  air  pressure  above  forces 
the  mercury  through  the  minute  pores  of  the  wood  into  the 
catch  basin  below.  This  demonstration  also  proves  that  the 
wood  is  porous. 

191.  Torricelli's    Experiment. — The    ancients    were    well 
aware  of  the.  fact  that  water  would  rise  against  gravity  in 


182 


MECHANICS 


exhausted  tubes.  This  knowledge  was  utilized  in  lifting 
water  from  wells.  The  accepted  explanation  was  that 
" nature  abhors  a  vacuum."  It  was  discovered  early  in  the 

seventeenth  century  that  water 
would  not  rise  much  over  32  ft. 
in  an  exhausted  tube.  Galileo's1 
explanation  of  this  phenomenon  was 
that  "  nature's  horror  of  a  vacuum 
ceased  at  32  ft."  It  is  probable, 
however,  that  Galileo  suspected  the 
true  reason  just  before  he  died. 
After  Galileo's  death,  Torricelli,2  his 
pupil,  reasoned  since  mercury  is 
about  13  times  as  heavy  as  water,, 
that  mercury  would  rise  only  about 
J^3  as  high  as  water  in  an  exhausted 
tube.  In  1643  Torricelli  performed 
the  following  experiment.  He  pro- 
cured a  glass  tube  about  4  ft.  long 
and  closed  at  one  end.  He  then 
filled  the  tube  with  mercury,  thereby  expelling  the  air  (Fig. 
197) .  Placing  his  finger  over  the  open  end,  he  inverted  the  tube 
in  a  bath  of  mercury.  The  mercury  then  fell  until  its  upper 
surface  was  about  30  in,  above  the  level  of  the  mercury  in  the 
dish,  proving  the  original  contention  that  mercury  would 
rise  Jis  as  high  as  water.  Torricelli  attributed  the  level  of 
the  mercury  in  the  tube  to  the  pressure  of  the  air. 

192.  Pascal's  Experiment. — Pascal  argued,  if  atmospheric 
pressure  were  the  supporting  force,  that  the  column  of  mercury 
would  fall  somewhat  at  higher  altitudes  where  the  air  is  rarer. 

1  Galileo  Galilei  (1564-1642).  Born  in  Italy.  Professor  of  mathe- 
matics at  Pisa  and  Padua.  Discovered  the  laws  of  falling  bodies  and  the 
pendulum.  Constructed  the  first  thermometer.  Made  many  valuable 
discoveries  in  astronomy.  Considered  the  father  of  modern  physics. 

2Evangelista  Torricelli  (1608-1647).  Italian  physicist  and  mathe- 
matician. Successor  of  Galileo.  Constructed  the  first  microscope  and 
improved  the  telescope. 


FIG.  197. — Torricelli' s  experi- 
ment. 


FLUIDS  183 

Accordingly,  he  carried  Torricelli's  apparatus  to  the  top  of 
a  tower  in  Paris  and  noted  that  there  was  a  perceptible  drop 
in  the  mercury  column.  A  Torricellian  apparatus  may  be 
used  to  determine  altitude.  At  places  near  the  sea  level,  the 
mercury  will  drop  about  1  mm.  for  every  11  meters  of  ascent 
or  J^o  of  an  in.  for  every  90  ft.  of  ascent. 

We  may  verify  Pascal's  experiment  by  the  fol- 
lowing demonstration.  Referring  to  Fig.  198, 
we  have  an  exhausted  tube  containing  mercury 
to  the  approximate  height  of  30  in.  or  760  mm., 
the  open  end  dipping  into  a  well  of  mercury. 
The  apparatus  is  placed  upon  the  plate  of  an  air 
pump  and  covered  with  a  high  bell  jar.  The 
air  is  then  exhausted  within  the  jar  and  the 
mercury  will  fall  until  it  is  almost  at  the  level  of 
the  mercury  in  the  well.  If  the  air  is  allowed  to 
re-enter  the  bell  glass,  the  mercury  will  rise  to  its 
former  position.  This  shows  conclusively  that 
the  rise  of  liquids  in  exhausted  tubes  is  due  to 
atmospheric  pressure. 

193.  The  Mercury  Barometer. — The  ordinary 
mercury  barometer  is  simply  a  Torricellian  tube 
mounted  as  shown  in  Fig.  199 (a).  The  tube 
terminates  at  the  lower  extremity  in  a  generous  FlG-  1 9 8~ 
sized  reservoir  with  a  flexible  bottom  (usually  prove'that  the 
chamois).  At  the  left,  the  scale  is  graduated  mercury  coi- 
in  inches  and  tenths  of  inches;  at  the  right,  Ionian  tubers 
the  corresponding  range  in  millimeters.  The  supported  by 
protecting  cap  over  the  reservoir  is  constructed  pr^ure.  G 
so  as  to  afford  a  clear  view  of  the  mercury  as  well 
as  the  fixed  white  glass  pointer  (see  Fig.  199(6))  which  should 
always  coincide  with  the  convex  surface  of  the  mercury. 
Before  reading  the  instrument,  the  thumbscrew  must  be  so 
regulated  that  the  pointer  and  the  surface  of  the  mercury  will 
just  meet.  If  the  pressure1  of  the  atmosphere  increases,  the 

1  The  rear  of  the  mercury  reservoir  has  a  porous  stopper. 


184 


MECHANICS 


mercury  will  ascend  in  the  tube;  if  the 
pressure  decreases,  it  will  descend  in  the 
tube.  In  reading  the  barometer,  great  care 
must  be  exercised  that  the  eye  and  the  con- 
vex surface  of  the  mercury  in  the  tube  are 
in  a  horizontal  line.  A  rising  barometer 
indicates  fair  weather;  a  falling  barometer 
indicates  that  a  storm  is  approaching.  A 
sudden  drop  at  sea  indicates  a  storm  of 
unusual  severity.  The  barometer  is  not  a 
sure  storm  prophet,  yet  it  is  very  helpful  in 
making  weather  predictions.  While  it  may 
not  storm  at  every  low  barometer,  we  know 
that  the  barometer  is  low  whenever  it  does 
storm.  The  mercury  barometer  is  used 
principally  for  stationary  work  and  where 
the  temperature  is  not  sufficient  to  freeze 
the  mercury.  Mercury  freezes  at  —  39°C. 


Pointer 


Mercury 
Reservoir 


Flexible 
'Bottom 


FIG.  199(6). — View  of  mercury  reservoir 
in  Fig.  199(a). 

194.  The  Aneroid  Barometer. — The  aner- 
oid barometer  is  shown  in  Fig.  200.  It 
consists  of  a  circular  box  of  metal,  the  sides 
of  which  are  thin,  elastic  and  corrugated. 
The  air  is  partially  exhausted  from  the 

FIG.  199(o). — Mer-  .    J  i   j       /\          -j 

cury  barometer,    box  and  the  box  is  then  sealed.    One  side 


FLUIDS 


185 


is  securely  fastened  to  the  base  and  the  other  side  oper- 
ates against  a  system  of  levers,  etc.,  which  in  turn  move 
the  pointer  around  the  graduated  dial.  If  the  outside  pressure 
increases,  the  front  face  of  the  box  moves  in,  registering  an 
increased  pressure  on  the  dial. 
If  the  outside  pressure  decreases, 
the  face  of  the  box  moves  outward 
and  the  pointer  moves  back  due 
to  the  spring  with  which  it  is 
connected.  Aneroid  barometers 
are  graduated  by  comparison  with 
a  standard  mercury  barometer. 
Aneroid  barometers  are  very  sen- 
sitive in  operation  but  they  fre- 
quently get  out  of  order  and  need 
re-graduating.  They  are  used  for 
work  in  which  it  is  necessary  to 
move  from  place  to  place  and  also  in  the  barograph  which 
will  be  described  in  the  next  paragraph. 

195.  The  Barograph. — The  barograph,  shown  in  Fig.  201, 
is  an  aneroid  instrument  designed  to  record  automatically 


FIG.  200. — Aneroid  barometer. 


FIG.  201. — Barograph  or  recording  barometer. 

the  varying  pressures  of  the  atmosphere.  The  drum,  revolv- 
ing once  every  seven  days,  carries  a  graduated  chart.  The 
variations  of  pressure  cause  movements  of  the  lever,  the  free 
end  of  which  carries  a  pen.  Thus  the  pressure  for  an  entire 
week  is  self-recorded  on  the  chart.  In  this  barometer  the 


186  MECHANICS 

motion  is  magnified  by  the  use  of  a  series  of  rarefied 
boxes  connected  one  to  another,  rather  than  a  single  box. 
The  barograph  may  be  seen  at  any  government  weather 
station. 

196.  Standard  Conditions  of  Pressure  and  Temperature.— 
Since  gas  volumes  are  affected  both  by  pressure  and  tempera- 
ture, it  necessarily  follows  that  experimenters  doing  similar 
quantitative  work  on  gases  in  different  localities  would  not 
agree  in  their  results,  due  to  different  conditions  of  pressure 
and  temperature.     In  order  to  obviate  this  difficulty,  standard 
conditions  of  pressure  and  temperature  have  been  adopted. 
Uniformity  is  obtained  by  reducing  all  numerical  work  to 
these  standards. 

Standard  conditions  for  gases  are  a  pressure  of  76  cm.  of 
mercury  and  a  temperature  of  0°C.  or  32°F.  For  ordinary 
calculations  either  76  cm.  or  30  in.  may  be  used. 

197.  The  Magdeburg  Hemispheres. — The  original  Magde- 
burg hemispheres,  invented  by  Otto  von  Guericke,1  consist  of 
two  hollow  metallic  hemispheres  fitting  tightly  together.     The 

spheres  are  22  in.  in  diameter  and  it  is 
said  that  four  teams  of  horses  on  either 
side  were  unable  to  separate  them  when 
the  air  was  exhausted,  due  to  the  pres- 
sure of  the  air  on  the  outer  surfaces. 

The  hemispheres  shown  in  Fig.  202  are 
used  for  ordinary  demonstrations  and  are 
4  in.  in  diameter.  The  lower  hemisphere 
terminates  in  a  threaded  stem  which  may 
be  removed  from  the  base  and  screwed 
into  the  plate  of  an  air  pump.  When  the 
'IG'  61311^  air  is  exhausted,  the  outside  pressure 

holding   them   together  is  so  great  that 
two  students  will  experience  great  difficulty  in  separating 

1  Otto  von  Guericke  (1602-1686).  German  physicist  and  astronomer. 
Mayor  of  Magdeburg.  Made  many  experiments  with  liquids  and 
gases.  Constructed  the  first  air  pump  in  1605.  Inventor  of  the  Magde- 
burg hemispheres  which  four  teams  of  horses  could  not  separate. 


FLUIDS  187 

them.  In  computing  the  total  pressure  tending  to  keep 
the  hemispheres  together,  simply  assume  a  circular  plane 
whose  diameter  is  the  same  as  the  diameter  of  the  hemi- 
spheres. The  curved  area  is  not  used  on  account  of  the 
fact  that  the  various  pressures,  acting  at  right  angles  to 
the  curved  surface,  do  not  all  point  in  the  same  direction  for 
each  individual  half.  The  components  of  these  pressures, 
acting  perpendicular  to  the  circular  plane  assumed,  comprise 
the  total  force  acting.  If  the  vacuum  is  perfect,  the  force 
necessary  to  pull  apart  the  hemispheres  shown  in  Fig.  202 
will  be  4  X  4  X  .7854  X  14.7  or  184.72  Ib. 

Questions  and  Problems 

1.  State  the  general  characteristics  of  gases. 

2.  What  pressure  is  exerted  by  the  atmosphere  at  sea  level? 

3.  What  is  the  average  density  of  air  at  sea  level  and  at  0°C.  ? 

4.  What  is  meant  by  the  height  of  the  homogeneous  atmosphere? 
6.  Describe  an  experiment  to  prove  that  air  has  weight. 

6.  Describe  an  experiment  to  prove  that  air  exerts  pressure. 

7.  Describe  Torricelli's  experiment. 

8.  Describe  Pascal's  experiment. 

9.  How  would  you  verify  Pascal's  experiment  in  the  laboratory? 

10.  Describe  the  construction  and  operation  of  a  mercury  barometer. 

11.  Describe  the  construction  and  operation  of  an  aneroid  barometer. 

12.  Is  the  barometer  a  good  weather  prophet?     Explain. 

13.  How  would  you  determine  the  altitude  by  means  of  a  barometer? 

14.  State  clearly  what  is  meant  by  standard  conditions  of  temperature 
and  pressure.     Why  were  standard  conditions  adopted? 

15.  Describe  the  original  Magdeburg  hemispheres.     How  would  you 
compute  the  force  necessary  to  separate  a  pair  of  such  hemispheres? 
Assume  a  vacuum  within. 

16.  What  is  the  weight  of  a  column  of  mercury  30  in.  high  and  1  sq. 
in.  in  cross-section?     What  pressure  does  the  column  exert? 

17.  What  is  the  weight  of  a  column  of  mercury  76  cm.  high  and  1 
sq.  cm.  in  cross-section?     What  pressure  does  the  column  exert? 

18.  A  water  barometer  reads  33  ft.     What  is  the  pressure  of  the  air? 

19.  Find  the  weight  of  the  air  in  a  room  20  X  30  X  12  ft.     Assume 
standard  conditions. 

20.  What  will  the  barometer  normally  read  at  the  top  of  the  Woolworth 
building  792  ft.  high? 

21.  If  the  atmosphere  had  the  same  density  throughout  as  at  the 


188 


MECHANICS 


earth's  surface,   how    high  would    it    extend    in  miles?     Assume  the 
height  of  the  barometer  to  be  30  in. 

22.  If  the  pressure  within  the  Magdeburg  hemispheres  (Prob.  15)  is 
sufficient  to  sustain  a  2-inch  column  of  mercury,  what  force  will  be  neces- 
sary to  separate  them? 

198.  Boyle's  Law.1 — Robert  Boyle2  was  the  first  man  to 
make  a  study  of  the  relation  of  a  confined  gas  and  the  pressure 

exerted  by  the  gas.     He  stated,  as 
a  result  of  his  investigations,  that, 
during  all  volume    and  pressure 
/J'A  y  changes  for  a  given  gas,  the  product 

of  the  volume  .and  the  pressure  will 
remain  constant.  For  example,  sup- 
pose that  we  have  4  cu.  ft.  of  air 
confined  under  a  pressure  of  1  atmo- 
sphere. The  product  of  the  volume 
and  pressure  will  be  4X1  or  4. 
Suppose,  further,  that  we  increase 
the  pressure  against  the  gas  to  2 
atmospheres.  The  volume  will  be 
reduced  to  2  cu.  ft.  and  the  product 
of  the  volume  and  pressure  will  be 
2  X  2  or  4.  The  above  relation  is 
usually  stated: 

The    temperature    remaining    con- 
FIG.  203.— Apparatus  for  veri-  slant,  the  volume  of  a  confined  gas 

fying  Boyle's  law.  .77  .  7  ., 

will  vary  inversely    as    the   pressure 
acting  upon  it. 

Boyle's  law  may  be  stated  algebraically: 

Vl  :  V2  =  P2  :  Pl  or  VlPl  =  V2P2. 

The  following  simple  experiment  will  verify  Boyle's  law. 
Figure  203  (a)  represents  a  glass  tube  of  uniform  bore,  bent  as 
shown  and  closed  at  C.  Mercury  is  poured  into  the  open  end 

1  Boyles  law  is  not  exact,  but  is  close  enough  for  practical  purposes. 

2  Robert  Boyle  (1627-1691).     English  physicist  and  chemist.     Experi- 
mented extensively  in  pneumatics. 


FLUIDS  189 

through  a  funnel  and  the  apparatus  adjusted  until  mercury 
levels  A  and  B  are  the  same.  The  volume  of  the  confined 
air  (proportional  to  BC)  is  then  under  a  pressure  of  1  atmos- 
phere or  76  cm.  of  mercury.  More  mercury  is  poured  into  the 
tube,  see  Fig.  203(6),  until  the  level  of  the  mercury  in  the 
open  part  is  76  cm.  above  the  level  in  the  closed  part.  The 
confined  gas  is  now  under  a -pressure  of  2  atmospheres  or  152 
cm.  of  mercury  and  the  volume  of  the  confined  air  (propor- 
tional to  C'B')  will  be  reduced  to  %  its  former  size.  By 
doubling  the  pressure  we  have  halved  the  volume. 

It  is  evident,  in  the  above  case,  that  the  density  of  the  air 
was  doubled  when  the  pressure  was  doubled.  Hence:  the 
density  of  a  confined  gas  varies  directly  as  the  pressure  upon  it. 
Algebraically : 

Dl  :D2  =  Pl  :P2. 

//  the  volume  remains  constant,  the  weight  of  a  confined  gas 
varies  directly  as  the  pressure  it  exerts.  Algebraically : 

W,:W2  =Pi:P2. 

Questions  and  Problems 

1.  State  Boyle's  law. 

2.  Describe  an  experiment  to  verify  Boyle's  law. 

3.  State  the  relation  between  the  density  of  a  confined  gas  and  the 
pressure  upon  it. 

4.  State  the  relation  between  the  weight  of  a  confined  gas  and  the 
pressure  exerted  by  the  gas,  volume  constant. 

5.  A  vessel  contains  5  liters  of  air  at  standard  pressure.     What  will 
the  volume  be  if  the  pressure  is  changed  to  (a)  780  mm.?     (6)  740  mm.? 

6.  10  cu.  ft.  of  illuminating  gas  is  at  a  barometer  pressure  of  30  in. 
What  will  be  the  pressure,  if  the  volume  is  compressed  to  7  cu.  ft.? 

7  The  gasoline  tank  of  an  automobile  lacks  10  gallons  of  being  full. 
The  engine  pump  communicating  with  the  tank  is  allowed  to  run  until 
the  air  gauge  on  the  dash  reads  2  Ib.     Find  the  weight  of  the  air  in  the 
tank.     1  cu.  ft.  of  air  at  standard  conditions  weighs  .0807  Ib. 

8  A  10  cu.  ft.  tank  of  helium  gas  (density  at  standard  conditions 
=  .0112  lb./cu.  ft.)  is  at  a  pressure  of  760  mm.  of  mercury.     Helium 

is  pumped  into  the  tank  until  the  pressure  is  50  Ib.  Find  (a)  the  density 
of  the  gas;  (6)  the  weight  of  the  gas. 

9.  The  weight  of  the  air  in  a  room  is  50  Ib.  at  29.5  in.  of  mercury. 
What  will  be  the  weight  at  30  in.  of  mercury? 


190  MECHANICS 

10.  A  1922  Buick  six  cylinder  motor  has  a  bore  of  3.375  in.  and  a 
stroke  of  4.5  in.     Assuming  the  clearance  to  be  1  in.,  what  pressure  will 
the  mixture  of  air  and  gasoline  vapor  exert  under  full  compression? 
What  will  be  the  total  pressure  against  the  cylinder  head?     Take  the 
atmospheric  pressure  as  15  Ib.     Assume  constant  temperature. 

11.  Repeat  as  above  for  an  eight  cylinder  Cadillac  motor.     The  bore 
is  3.125  in.  and  the  stroke  is  5.125  in. 

199.  The  Aeroplane. — Since  the  aeroplane  is  heavier  than 
air,  the  lifting  force  is  not  due  to  the  same  reason  as  in  the  case 
of  a  balloon.  The  lifting  force  is  due  to  the  reaction  or  thrust 
of  the  air  against  the  planes  as  the  machine  is  driven  through 


FIG.  204. — Curtiss  model  18-T  triplane. 

the  air  at  a  high  rate  of  speed.  The  velocity  is  obtained  by 
means  of  a  special  wood  propeller  driven  by  a  gasoline  motor 
of  special  design.  Aeroplanes  have  proven  indispensable  in 
war  for  scouting,  bombing,  etc.  They  have  yet  to  prove  their 
success  as  heavy  cargo  carriers. 

The  aeroplane  shown  in  the  above  cut  is  a  Curtiss  Model 
18-T  Triplane.  The  over-all  length  is  about  23  ft.  and  the 
over-all  height  about  10  ft.  The  wing  span  is  nearly  32  ft. 
The  machine  weighs  1,825  Ib.  and  will  carry  a  gross  weight  of 
2,901  Ib.  It  has  a  maximum  horizontal  flight  velocity  of 
163  m.p.h.  and  a  climbing  speed  of  15,000  ft.  in  10  minutes. 
At  2,500  r.p.m.,  the  motor  horsepower  is  rated  at  400. 


FLUIDS  191 

200.  The  Balloon. — The  lifting  ability  of  a  balloon  depends 
upon  the  buoyancy  of  the  air.  Just  as  a  body  submerged  in 
water  is  buoyed  up  by  the  weight  of  the  water  displaced,  so  a 
body  submerged  in  air  is  buoyed  by  the  weight  of  the  air  dis- 
placed. The  buoyant  force  of  water  is  62.5  Ib.  for  each  cubic 
foot  displaced.  Since  air  is  about  1/773  as  heavy  as  water, 
the  buoyant  force  of  a  cubic  foot  of  air  will  be  62.5/773  or 
approximately  .08  Ib.  A  balloon  containing  a  vacuum  would 
have  a  tremendous  lift,  but  would  be  crushed  by  the  outside 


FIG.  205. — U.  S.  Army  type  AC  dirigible. 

pressure.  To  obtain  a  suitable  lifting  force,  the  balloon  is 
filled  with  some  very  light  gas  as  hydrogen,  helium,  etc. 
Helium  is  today  regarded  as  the  most  satisfactory  gas.  It  is 
heavier  than  hydrogen,  but  is  not  inflammable  like  hydrogen. 
The  gas  densities  given  below  are  at  standard  conditions,  i.e., 
0°C.  and  76  cm.  of  mercury: 

Hydrogen 0056  Ib./cu.  ft. 

Helium 0112  Ib./cu.  ft. 

Illuminating  gas 0500  Ib./cu  ft. 

Air 0807  Ib./cu.  ft. 

The  lift  exerted  by  a  balloon  depends  upon  the  density  of 
the  gas  with  which  the  balloon  is  filled.  It  is  equal  to  the 
weight  of  the  air  displaced  by  the  envelope  minus  the  weight 
of  the  gas  in  the  envelope.  The  ascensional  force  is  the  force 


192  MECHANICS 

with  which  the  balloon  leaves  the  ground.  It  is  equal  to  the 
lift  minus  the  weight  of  the  envelope,  rigging,  car,  passengers, 
etc. 

Figure  205  shows  a  type  AC  dirigible  recently  built  for  the 
U.  S.  Army.  It  is  167  ft.  long,  has  a  gas  capacity  of  187,000 
cu.  ft.  and  a  cruising  speed  of  about  65  m.p.h.  The  car  is 
entirely  closed  and  will  carry  8  people. 

An  Army  observation  balloon  of  the  R  type  is  90  ft.  long,  29  ft.  in 
diameter  and  has  a  gas  capacity  of  37,000  cu.  ft.  The  envelope,  rigging, 
car,  etc.,  weigh  1,000  Ib. 

Suppose  we  wish  to  find  the  total  lifting  effect  and  the  ascensional  force 
at  ground  conditions,  if  the  envelope  is  filled  with  pure  hydrogen. 

Solution: 

.0807  -  .0056  =  .0751   Ib.    lifting   effect   per  cu.  ft.   of  hydrogen, 

37,000  X  .0751  =2,778.7  Ib.  total  lifting  effect, 

2,778.7  —  1,000  =  1,778. 7  Ib.  ascensional  force. 

Suppose  now  that  we  wish  to  find  the  ascensional  force  of  the  above 
balloon  if  filled  with  commercial  hydrogen  (98  per  cent.  pure).  Assume 
the  2  per  cent,  impurity  in  the  hydrogen  to  be  air. 

Solution: 

37,000  X  .98  =  36,260  cu.  ft.  of  pure  hydrogen, 

36,260  X  .0751  =  2,723.1  Ib.  total  lifting  effect, 

2,723.1  -  1,000  =  1,723.1  Ib.  ascensional  force. 

Questions  and  Problems 

1.  Explain  the  lifting  effect  of  an  aeroplane. 

2.  What  is  meant  by  the  lift  and  ascensional  force  of  a  ballon? 

3.  State  how  each  of  the  above  is  determined. 

4.  State  the  various  gases  used  in  ballons.     What  is  the  advantage 
of  each? 

5.  Why  does  a  submarine  either  sink  to  the  bottom  or  float  on  the 
surface  when  the  power  is  cut  off,  while  a  balloon,  on  ascending,  will 
reach  a  certain  height  and  float? 

6.  A  spherical  ballon  50  ft.  in  diameter  is  filled  with  pure  hydrogen. 
Find:  (a)  the  buoyant  force  of  the  air;  (6)  the  weight  of  the  hydrogen; 
(c)  the  total  lift. 

7.  Repeat  Problem  6,  using  pure  helium. 

8.  Repeat  Problem  6,  using  98  per  cent,  pure  hydrogen. 

9.  Repeat  Problem  6,  using  92  per  cent,  pure  helium. 

10.  A  spherical  balloon  has  a  capacity  of  40,000  cu.  ft.  and  is  filled  with 
commercial  hydrogen  (98  per  cent.  pure).  The  balloon  and  car  weigh 


FLUIDS 


193 


450  Ib.     Two  passengers  weighing  160  Ib.  each  are  carried.     Find:  (a) 
the  total  lift;  (6)  the  ascensional  force. 

11.  An  84,000  cu.  ft.  U.  S.  Army  dirigible  is  filled  with  pure  hydrogen. 
The  gas  bag,  car,  motors,  crew,  etc.,  weigh  6,300  Ib.  What  is  the 
ascensional  force? 

201.  The  Siphon. — The  siphon  is  a  device  used  to  transfer 
liquids  from  one  receptacle  to  another.     It  consists  of  a  bent 
tube   of  unequal  arms.     A  rubber  tube  may   be   used  for 
ordinary  liquids.     Referring  to  Fig. 

206,  let  us  study  the  operation  of 
a  siphon.  Assume  that  water  is 
flowing  from  vessel  A  into  vessel 
B  through  the  bent  tube.  Since 
EF  is  shorter  than  CG,  it  is  evident 
that  the  weight  of  the  liquid  in  the 
latter  arm  is  greater  than  in  the 
former.  The  upward  pressure  at  F 
is  the  atmospheric  pressure  minus 
the  pressure  of  the  column  EF.  The 
upward  pressure  at  G  is  the  atmos- 
pheric pressure  minus  the  pressure 
of  the  column  CG.  Therefore  the 
upward  pressure  at  F  exceeds  the  upward  pressure  at  G. 
It  is  evident  that  there  will  be  a  flow  from  the  point  of 
greater  pressure  at  F  to  the  point  of  less  pressure  at  G. 
The  flow  will  continue  until  F  and  G  are  at  the  same  level  and 
hence  at  the  same  pressure.  If  EF  is  higher  than  34  ft.,  the 
siphon  will  not  work,  as  water  will  not  rise  above  that  point 
in  an  exhausted  tube.  If  mercury  is  used,  EF  must  not 
exceed  30  in. 

202.  The  Air  Pump. — The  so-called  air  pump  is  an  appa- 
ratus for  exhausting  air  from  vessels.     It  was  invented  in 
1650  by  Otto  von  Guericke.     With  a  good  pump,  a  very  high 
degree  of  vacuum  may  be  obtained.     Figure  207  shows  the 
construction  of  a  simple  air  pump.     It  consists  of  a  cylinder 
(C)  with  a  tight-fitting  piston  (P).     Vi  and  V2  are  valves. 
On  the  upward  stroke,  a  vacuum  is  formed  under  the  piston 


t 

F 

'-_ 

Y 

N, 

~- 

I 

_6_ 

~~  ~. 

-_ 

~-~~ 



-- 

Wafer 

"Wafer" 

A  B 

FIG.  206.— Siphon. 


194 


MECHANICS 


in  the  cylinder,  V\  closes  automatically  and  Vi  opens,  and  the 
air  in  the  bell  glass  (B)  passes  in  part  to  the  cylinder  to  fill  up 
the  vacuum  and  equalize  the  pressures.  On  the  downward 
stroke,  F2  closes  and  V\  opens,  allowing  the  air  in  the  cylinder 
to  pass  out.  The  above  process  is  repeated,  each  time  a 


FIG.  207. — Air  pump. 

portion  of  the  air  in  B  being  removed.  Since  only  a  fraction 
of  the  air  is  removed  at  each  stroke,  it  will  be  seen  that  a 
perfect  vacuum  can  never  be  obtained.  The  air  pump  may 
be  converted  into  a  compression  pump  by  opening  F2  to  the 
atmosphere  and  attaching  the  receptacle,  into  which  the  air  is 
to  be  compressed,  to  the  tube  at  V\. 

203.  The  Air  Brake. — Figure  208  shows  the  construction  of  a 

simple  Westinghouse  air 
brake.  T  is  a  tank  car- 
ried by  each  separate  unit, 
in  which  the  engine  pump 
maintains,  through  pipe 
(P),  a  pressure  of  about 
75  Ib.  While  the  engine 
pressure  is  acting  in  P, 
the  triple  valve  V  main- 
tains communication 


FIG.  208. — Westinghouse  air  brake. 


between  P  and  T.     If  the  pressure  is  cut  off  in  P,   either 
at  the  engine  or  by  accidental  uncoupling  of  the  pipe  line, 


FLUIDS 


195 


communication  between  P  and  T  is  cut  off  at  V  and  communi- 
cation is  opened  between  the  tank  (T)  and  the  cylinder  (C). 
Thus  the  compressed  air  in  T  forces  the  piston  to  the  left  in 
the  cylinder,  setting  the  brake  shoes  against  the  wheels.  To 
remove  the  friction  of  the  brakes,  pressure  is  again  turned  on 
in  P.  The  connection  between  T  and  C  is  thus  closed,  the  air 
from  C  exhausts  at  E  and  the  spring  (S)  moves  the  piston 
back  to  its  original  position. 

204.  Liquid   Pumps   with   Reciprocating   Parts. — Recipro- 


~~WafeF 
FIG.  209. — Lift  pump. 


Y/afer 
FIG.  210. — Force  pump. 


eating  pumps  ordinarily  depend  upon  atmospheric  pressure 
for  their  operation.  The  lift  pump  (Fig.  209)  consists  of  a 
cylinder  with  a  tight-fitting  piston.  The  piston  carries  a 
valve  (V)  and  the  lower  part  of  the  cylinder  has  a  valve  (Fi). 
Whenever  the  piston  is  drawn  up,  V  closes  and  Vi  opens,  water 
being  forced  up  into  the  cylinder  by  the  pressure  of  the  air. 
On  the  down  stroke  V\  closes,  trapping  the  water  already  in  the 
cylinder.  At  the  same  time  V  opens,  allowing  the  trapped 
water  to  pass  into  the  upper  part  of  the  cylinder.  On  the 
next  up  stroke  V  closes,  trapping  the  water  above  the  piston 
and  forcing  it  out  for  delivery.  Simultaneously  Vi  opens  and 
the  previous  cycle  is  repeated.  The  pump  will  not  operate  if 


196  MECHANICS 

Vi  is  much  more  than  30  ft.  above  the  surface  of  the  supply 
water. 

Figure  210  shows  a  force  pump.  Vi  opens  as  the  piston  is 
drawn  up,  and  water  is  forced  up  into  the  cylinder  by  atmos- 
pheric pressure;  at  the  same  time  V  closes.  On  the  down 
stroke  Vi  closes,  trapping  the  water  in  the  cylinder,  and  V 
opens,  allowing  the  water  to  pass  into  the  delivery  tube. 


FIG.  211. — Goulds  "Triplex"  water  pump. 

On  the  next  up  stroke  V  closes,  V\  opens  and  the  cycle  is 
repeated.  The  air  chamber  at  the  top  of  the  delivery  tube 
causes  the  water  flow  from  the  orifice  in  a  steady  stream, 
instead  of  intermittently.  There  is  no  reasonable  limit  to  the 
height  that  the  water  may  be  forced,  provided  that  Vi  is 
not  too  far  above  the  water  supply. 

Figure  211  is  from  a  photograph  of  a  Goulds  single-acting 
"Triplex"  pump.  This  type  of  pump  is  very  widely  used. 
It  has  a  maximum  working  pressure  of  300  Ib.  At  this  pres- 
sure, it  will  elevate  water  to  a  height  of  nearly  700  ft.  Atten- 
tion is  called  to  the  large  air  chamber  in  front. 


FLUIDS 


197 


205.  Liquid  Pumps  with  Rotating  Parts. — Pumps  with 
rotating  parts  have  come  into  great  prominence  during  the 
past  few  years.  Figure  212  shows  a  gear  pump.  Its  opera- 
tion will  be  clear  from  an  examination  of  the  figure.  This  type 


FIG.  212. — Gear  pump. 


FIG.  213 — Exposed  section 
of  a  centrifugal  pump. 


FIG.  214. — Worthington  centrifugal  fire  pump. 

of  pump  is  used  when  the  volume  of  liquid  to  be  pumped  is 
small.  Many  automobiles  make  use  of  gear  pumps  to  actuate 
the  oil  circulation  in  the  motor. 

Centrifugal  pumps  have  been  used  for  some  time  for  low 
pressure  purposes,  such  as  circulating  the  water  in  internal 
combustion  engines.  They  are  now  accepted  as  the  proper 
kind  of  pumps  for  delivering  liquids  against  high  heads. 


198  MECHANICS 

Those  designed  for  high  pressure  are  either  electrically  or 
steam  turbine  driven.  In  principle  the  centrifugal  pump  is  a 
reaction  turbine  working  backwards.  Its  construction  will 
be  seen  from  Fig.  213.  The  liquid  enters  at  the  center  (shown 
by  the  dotted  circle).  As  it  strikes  the  blades,  it  is  thrown 
toward  the  circumference  of  the  casing.  The  liquid  passes 
into  the  delivery  pipe  (D)  with  a  pressure  depending  upon  the 
design  and  speed  of  the  pump. 

Figure  214  shows  a  Worthington  centrifugal  pump  used  for 
fire  purposes.  It  is  turbine  driven  and  will  furnish  1,500 
gallons  of  water  per  minute  against  a  pressure  of  100  Ib. 

206.  Instruments  for  Measuring  Pressure.— The  commer- 
cial form  of  pressure  gauge  used  on  steam  boilers,  gas 


FIG.  215. — Pressure  gauge.  FIG.  216. — Vacuum  gauge. 

tanks,  etc.  (Fig.  215)  is  graduated  in  Ib./sq.  in.  This  type  of 
gauge  is  designed  to  register  the  pressure  in  excess  of  atmos- 
pheric. Thus,  if  the  gauge  reads  100  Ib./sq.  in.,  the  actual 
pressure  of  the  gas  under  test  is  115  Ib./sq.  in.  The  vacuum 
gauge  (Fig.  216)  is  used  to  measure  the  degree  of  rarefaction 
in  a  closed  gas  container  and  is  graduated  in  inches.  Zero 
in.  corresponds  to  a  pressure  of  15  Ib.  and  30  in.  corre- 
sponds to  a  vacuum  or  zero  Ib.  Hence,  if  the  gauge  reads 
15  in.,  the  actual  pressure  is  15/30  of  15  or  7.5  Ib. 

Figure  217  shows  a  laboratory  model  to  illustrate  the  pres- 
sure and  vacuum  gauge.  The  hollow  curved  tube  communi- 
cates with  the  gas  under  test.  Whenever  there  is  an  increase 


FLUIDS  199 

in  pressure,  the  tube  tends  to  straighten,  as  a  piece  of  garden 
hose  when  the  water  is  turned  on.  The  free  end  of  the  tube 
moves  out  and  communicates  its  motion  to  a  lever,  the  teeth 
on  the  extremity  of  which  mesh  with  a 
pinion  carrying  the  indicator.  If  the  pres- 
sure decreases,  the  tube  will  bend  in,  allow- 
ing the  indicator  to  move  in  the  opposite 
direction,  due  to  the  spirally  wound  spring 
with  which  the  indicator  is  connected. 
The  indicator  will  remain  stationary  when 
a  vacuum  has  been  obtained. 

The  open-arm  manometer  shown  in  Fig. 
218,  may  be  used  for  pressures  above   or 
below  atmospheric.     It  consists  of  a  tube 
bent  as  shown,   open  at  A,   containing  a      FIG.  217.— Work- 
liquid  of  known  density  and  connected  to   ing  parts  of  a  pres" 

sure  gauge. 

the  gas  container  at  C.     If  the  pressure  in 

the   container  is   greater   than    atmospheric,    the   liquid   in 

arm  A  will  rise  above  the  level  of  the  liquid  in  arm  B. 
The  height  of  the  liquid  in  A  over  the 
height  in  B  determines  the  excess  of  the 
pressure  in  the  container  over  atmos- 
pheric. The  atmospheric  pressure  plus 
the  excess  pressure  indicated  gives  the 
actual  pressure  exerted  by  the  gas.  If 
the  confined  gas  is  under  a  pressure  of 
less  than  one  atmosphere,  the  liquid  in  B 
will  rise  above  the  level  in  A .  The  excess 
of  the  atmospheric  over  the  actual  pres- 
sure exerted  by  the  gas  under  test  is  com- 
puted from  the  difference  in  height  of 
the  liquid  columns.  The  actual  pressure 
FIG.  218. — Open-arm  exerted  will  be  the  atmospheric  pressure, 
as  determined  by  the  barometer,  minus 

the  pressure  as  indicated  by  the  liquid  columns. 

The  closed-arm  manometer  shown  in  Fig.  219  is  used  to 

measure    higher    pressures    than    the  open-arm  manometer. 


200 


MECHANICS 


FIG.  219.— Closed-arm 
manometer. 


The  tube  is  connected  at  C,  as  in  the  previous  case,  but  arm  A 
is  closed.  The  liquid  used  is  mercury  and  the  space  above 
the  mercury  in  the  closed  arm  is  filled 
with  air.  The  pressure  is  determined 
by  means  of  the  change  in  the  height  of 
the  mercury  columns  and  by  the  change 
in  the  air  volume.  At  ordinary  atmos- 
pheric pressure  the  mercury  levels  will  be 
the  same.  Before  C  is  connected  to  the 
confined  gas,  the  position  of  the  mercury 
levels  is  noted  as  well  as  the  volume 
of  the  air  in  am  A.  The  barometer 
reading  is  also  taken.  As  the  arm  is  uni- 
form in  bore,  the  volume  is  proportional 
to  the  length  of  the  air  column.  Length 
may  therefore  be  used  instead  of  volume. 
The  pressure  is  computed  from  the 
change  in  the  mercury  levels  and  from  the  change  in  the 
length  of  the  air  column  according  to  Boyle's  law. 

Questions  and  Problems 

1.  Explain  the  action  of  the  siphon. 

2.  Describe  the  construction  and  operation  of  an  air  pump.     How 
may  the  air  pump  be  converted  into  a  compression  pump? 

3.  Why  is  it  impossible  to  obtain  a  perfect  vacuum  with  an  air  pump? 

4.  Describe  the  construction  and  operation  of  the  Westinghouse  air 
brake. 

6.  Into  what  two  classes  may  liquid  pumps  be  divided?     Give  the 
construction  and  operation  of  the  liquid  pumps  described  in  this  chapter. 

6.  What  is  the  difference  between  a  pressure  gauge  and  a  vacuum 
gauge? 

7.  Describe  the  principle  upon  which  the  commercial  gauge  works. 

8.  Describe  how  pressure  is  determined  with  an  open-arm  manometer. 

9.  Describe  how  pressure  is  determined  with  a  closed-arm  manometer. 

10.  If  in  Fig.  206  EF  is  20  in.  and  CG  is  40  in.,  compute  the  upward 
pressure  at  F  and  the  upward  pressure  at  G.     Barometer  reads  30  in. 

11.  Assuming  the  diameter  of  C  in  Fig.  208  to  be  12  in.  and  the  pres- 
sure in  T  to  be  75  Ib./sq.  in.,  what  is  the  total  pressure  against  the 
piston,  if  connection  between  T  and  C  is  open?     State  two  reasons  why 
the  piston  rod  will  not  transmit  the  entire  amount  of  the  total  pressure 


FLUIDS  201 

12.  If  the  reading  of  a  pressure  gauge  is  25  Ib./sq.  in.,  what  is  the  total 
pressure  exerted  by  the  gas  under  test? 

13.  If  a  vacuum  gauge  reads  20  in.,  what  is  the  actual  pressure  of  the 
gas  under  test? 

14.  The  open-arm  manometer  (Fig.  218)  is  connected  at  C  to  a  gas 
tank  through  a  two-way  valve.     When  the  valve  is  opened  to  the  air, 
the  water  levels  are  the  same.     When  the  valve  is  opened  to  the  gas 
tank,  the  level  in  A  is  3  in.  above  the  level  in  B.     Find  the  actual  gas 
pressure  in  the  tank. 

16.  Repeat  Problem  14,  if  the  level  in  B  is  2  in.  higher  than  in  A. 

16.  Find  the  pressure  of  the  confined  gas  in  problem  14,  if  the  level  in 
A  is  20  mm.  higher  than  the  level  in  B.     Give  answer  in  Kg. /cm2. 

17.  Before  connection  to  a  gas  tank,  the  level  of  mercury  in  the  closed 
arm  of  a  manometer  containing  mercury  is  2  cm.  higher  than  in   the 
opposite  arm.     After  connection  to  the  gas  tank,  the  difference  in  mercury 
levels  is  10  cm.  and  the  air  column  is  only  half  as  long.    What  is  the  actual 
pressure  of  the  gas  in  the  tank? 


CHAPTER  XXII 


t 


FALLING  BODIES;  CENTRIFUGAL  FORCE;  THE 
PENDULUM 

SECTION  1.     FALLING  BODIES 

207.  Effect  of  Gravity  on  Falling  Bodies. — Peculiar  as  it  may 
seem  to  the  student,  gravity  acts  with  equal  effect  on  all 
bodies.  If  a  heavy  body  and  a  light  body  are  dropped  from 
the  same  height  at  the  same  time,  they  will  reach  the  ground 
simultaneously,  provided  that  we  neglect  the 
resistance  of  the  air.  The  air  exerts  a  retard- 
ing effect  on  falling  bodies,  which  depends 
upon  the  surface  exposed.  For  example,  a 
piece  of  aluminum  weighing  5  Ib.  will  fall 
somewhat  more  slowly  than  a  piece  of  gold 
weighing  5  Ib.,  due  to  the  fact  that  the  alumi- 
num offers  more  area  to  the  air.  In  a  vac- 
uum, however,  they  will  fall  with  identical 
speeds.  This  may  be  proved  as  follows. 
Figure  220  represents  a  glass  tube  4  ft.  long 
and  closed  at  one  end.  The  lower  end  termi- 
nates in  a  petcock  with  an  air  pump  attach- 
ment. Within  the  tube  is  a  feather  and  a 
penny.  If  the  tube  is  quickly  inverted,  the 
ratuT  to  show  that  Penny  will  be  seen  to  fall  more  rapidly  than 
at  the  feather.  If  the  air  is  then  withdrawn 
a  from  the  tube  and  the  experiment  repeated, 
the  feather  and  penny  will  be  seen  to  fall 
with  equal  velocities.  Hence  we  may  arrive  at  the  conclusion 
that  bodies,  falling  from  the  same  height  in  a  vacuum,  will  fall 
equal  distances  in  equal  lengths  of  time.  As  air  resistance  is 
difficult  to  determine,  it  will  not  be  taken  into  account  in 
the  solution  of  problems. 

202 


all 


bodies    fall 
rate 


FALLING  BODIES 


203 


208.  The  Acceleration  of  Gravity. — It  has  been  determined 
by  experiment  that  the  acceleration  of  gravity  is  constant; 
that  is,  a  freely  falling  body  will  be  imparted  an  equal  gain  in 
velocity  for  each  succeeding  second.     Starting  with  a  velocity 
of  zero,  the  body  will  have  a  velocity  of  32.16  ft./sec.  at  the 
end  of  the  first  second;  64.32  ft./sec.  at  the  end  of  the  second 
second,  etc.     Thus  we  have  an  example  of  uniformly  acceler- 
ated motion.     The  force  of  gravity  varies  slightly  from  place 
to    place.     At    New    York,    the    acceleration    of    gravity 
(symbol  "gr")  is  about  32.16  ft.  per  second  per  second  or  980 
cm.  per  second  per  second.     It  is  usually  written  32.16  ft./ 
sec.2  or  980  cm. /sec.2     During  the  first  second,  a  freely  falling 
body  will  travel  16.08  ft.  or  490  cm. 

209.  Freely  Falling  Bodies. — We  may  now  construct  the 
following  table  for  freely  falling  bodies,  allowing  D  to  repre- 
sent %  g  or  the  distance  passed  over  during  the  first  second. 
Since  the  acceleration  is  constant,  the  body  will  gain  2D 
in  velocity  each  second  and  will  pass  over  during  each  succeed- 
ing second  2D  more  than  in  the  previous  second. 


/ 

V 

* 

S 

1 

2D 

D 

D 

2 

4D 

3D 

4D 

3 

QD 

5D 

9D 

4 

SD 

7D 

16D 

in  which 


t  =  the  time  in  seconds, 

v  =  the  velocity  at  the  end  of  any 

particular  second, 
s  =  the  distance  passed  over  dur- 
ing any  particular  second, 
S  =  the  total  distance  passed  over 
at  the  end  of  a  given  number 
of  seconds. 


Substituting  ^  g  for  D,  it  is  evident  that  : 

1.  v   =  gt 

2.  s   =  Y2g(2t  -  1) 

3.  S  = 


210.  Bodies  Rolling  Down  an  Incline.  —  The  acceleration 
for  bodies  rolling  down  an  incline  will  be  less  than  g.  It  is 
found  by  multiplying  g  by  the  height  (h)  of  the  plane  over  the 
length  (I)  of  the  plane. 

Acceleration  on  an  incline  =  g  X  h/l 


204  MECHANICS 

Since  h/l  is  equal  to  the  sine  of  the  angle  of  inclination,  the 
following  formula  may  be  used : 

Acceleration  on  an  incline  =  g  sin  6 

211.  Bodies  Projected  Vertically  Downward. — In  case  a 
body  is  projected  vertically  downward  from  a  state  of  rest 
(e.g.,  a  ball  thrown  down  from  a  high  building),  the  formulas 
given  in  the  last  paragraph  must  be  changed  to  read : 

1.  v   =•  Initial  velocity  +  gt 

2.  s  =  Initial  velocity  +  ^g(2t  —  1) 

3.  S  =  Initial  velocity  X  t  +  %gt* 

212.  Bodies  Projected  Vertically  Upward. — If  a  body  is 
projected  vertically  upward,  the  acceleration  of  gravity  is 
negative;  i.e.,  the  body  will  lose  32.16  ft./sec.  or  980  cm. /sec. 
until  a  velocity  of  zero  is  obtained.     Starting  from  zero  the 
body  will  return,  the  time  of  ascent  and  descent  being  equal, 
as  well  as  the  initial  and  final  velocity.     The  time  of  ascent 
(0  =  v/g. 

Questions  and  Problems 

(Assume  g  =  32  ft./sec.2  or  980  cm. /sec.2) 

1.  Describe  an  experiment  to  show  that  all  bodies  fall  at  the  same 
rate  in  a  vacuum. 

2.  Explain  what  is  meant  by  the  acceleration  of  gravity. 

3.  Find  v  and  S  for  a  body  freely  falling  from  rest  for  6  sec. 

4.  A  body  is  rolling  down  a  frictionless  inclined  plane  20  ft.  long  and 
5  ft.  high.     Give  the  acceleration  in  feet  and  centimeters. 

6.  A  ball  is  dropped  from  the  Woolworth  Building  at  a  point  600  ft. 
above  the  sidewalk.  Find  the  time  of  descent  and  the  velocity  on  strik- 
ing the  walk. 

6.  In  problem  5,  how  far  did  the  ball  travel  in  the  fourth  second? 
In  the  fourth  and  fifth  seconds  together? 

7.  How  long  will  it  take  a  spherical  body  to  roll  down  a  10  per  cent, 
grade,  if  the  grade  is  100  meters  long? 

8.  What  distance  did  the  body  in  the  previous  problem  pass  over 
during  the  second  second?     What  was  the  final  velocity? 

9.  A  stone  is  thrown  vertically  downward  from  a  height  of  200  ft. 
with  an  initial  velocity  of  10  ft./sec.     (a)  With  what  velocity  did  it 
strike  the  ground?     (6)  How  long  was  it  in  the  air?     (c)  How  far  did  it 
travel  in  the  second  second? 


FALLING  BODIES 


205 


10.  A  body  is  projected  vertically  upward  with  a  velocity  of  100 
meters  per  sec.     (a)  How  long  was  it  in  rising?     (6)  How  high  did  it 
rise?     (c)  How  long  was  it  in  the  air?     (d)  How  far  did  it  travel  in  the 
fifth  sec.?     (e)  With  that  velocity  did  it  strike  the  ground. 

11.  A  golf  ball  is  dropped  from  a  height  of  200  ft.  and  %  sec.  later  a 
second  ball  is  dropped.     How  far  apart  will  they  be  when  the  first  ball 
strikes  the  ground? 

213.  Bodies  Projected  Horizontally. — If  a  body  is  dropped 
from  a  certain  point  and  another  body  is  projected  horizontally 
from  the  same  height  at  the  same  time,  both  bodies  will  reach 


75 


B 

FIG.  221. — AEFC  represents  the  path  of  a  body  projected  horizontally  from  A. 

the  ground  at  the  same  instant.  The  second  body  will  advance 
horizontally  at  a  uniform  rate  of  speed,  but  will  curve  down- 
ward in  a  parabolic  path  due  to  the  action  of  gravity. 

AEFC  (Fig.  221)  represents  the  path  of  a  body  projected 
horizontally  from  A  with  a  velocity  of  75  ft. /sec.  For  con- 
venience we  will  assume  that  A  is  144.72  ft.  above  the  earth, 
so  that  the  time  in  the  air  will  be  exactly  three  seconds.  At 
the  end  of  the  first  sec.,  the  body  will  have  advanced  horizon- 
ally  a  distance  of  75  ft.  and  gravity  will  have  drawn  it  16.08 
ft.  toward  the  earth.  Therefore,  at  the  end  of  the  first 
second,  the  body  will  be  at  E.  At  the  end  of  the  second 
second  the  body  will  have  advanced  75  ft.  more  horizontally 
and  gravity  will  have  drawn  it  48.24  ft.  nearer  the  earth  or 
64.32  ft.  below  the  line  AD.  Hence,  at  the  end  of  the  second 


206 


MECHANICS 


second,  the  body  will  be  at  F.  Similarly,  it  can  be  shown  that 
the  body  will  be  at  C  at  the  end  of  the  third  second.  The 
distance  BC  is  called  the  range.  The  actual  path  of  the  body 
is  called  the  trajectory. 

Suppose  we  have  a  body  projected  horizontally  from  a  height  of 
257.28  ft  with  a  velocity  of  1,000  ft./sec.  It  is  required  to  find:  (1) 
the  time  of  descent;  (2)  the  vertical  velocity  on  striking;  (3)  the  range. 

Solution: 

1.  S  =  1/2  gt,2  257.28  =  16.08  *2,  t*  =  16,  t  =  4  sec. 

2.  v  =  at,  v  =  32.16  X  4  =128.64  ft./sec. 

3.  1,000  X  4  =  4,000  ft. 

214.  Bodies  Projected  so  that  the  Angle  of  Elevation  is 
Less  than  90°. — Suppose  a  body,  projected  with  a  velocity 


FIG.  222. — ARK  represents  the  path  of  a  body  projected  from  A  at  an  angle 
of  40°  with  the  horizontal. 

of  200  ft./sec.  from  A,  makes  an  angle  of  40°  with  the  ground 
(Fig.  222).  Were  it  not  for  gravity,  the  body  would  continue 
at  the  same  rate  indefinitely  in  the  straight  line  AG.  On 
account  of  gravity,  the  body  will  be  16.08  ft.  below  B  at  the 
end  of  the  first  sec.,  64.32  ft.  below  C  at  the  end  of  the  second 
second;  144.72  ft.  below  D  at  the  end  of  the  third  sec.,  etc. 
The  actual  path  of  the  body  will  be  parabolical. 


FALLING  BODIES  207 

To  solve  problems  in  the  above  case,  it  is  necessary  to  resolve  the 
initial  velocity  into  vertical  and  horizontal  components  (F  and  H)  and 
consider  them  as  separate  velocities.  Using  data  as  in  the  previous 
paragraph,  suppose  we  wish  to  find:  (1)  the  time  of  ascent;  (2)  the  time 
in  the  air;  (3)  the  final  vertical  velocity;  (4)  the  greatest  vertical  height 
(A);  and  (5)  the  range. 

Solution: 

First  it  is  necessary  to  resolve  200  ft. /sec.  into  its  vertical  and  hori- 
zontal components.  By  trigonometry: 

V  =  200  X  cos  50°,  V  =  200  X  .643  =  128.60  ft./sec. 
H  =  200  X  cos  40°,  H  =  200  X  .766  =  153.20  ft./sec. 

We  may  now  consider  the  body  to  have  an  initial  vertical  velocity  of 
128.60  ft./sec.  and  a  uniform  horizontal  velocity  of  153.20  ft./sec. 

1.  t  =  v/g,  t  =  128.60/32.16  =  4  sec. 

2.  4  X  2  =  8  sec. 

3.  Final  vertical  velocity  =  initial  vertical  velocity  =  128.60  ft./sec. 

4.  S  =  K  Qt,z  S  =  16.08  X  16  =  257.28  ft. 

5.  153.2  X  8  =  1,225.60  ft. 

Problems 

(Assume  g  =  32  ft./sec.2  or  980  cm. /sec.2) 

1.  Construct  a  graph  showing  the  path  of  a  body  projected  horizontally 
from  a  point  400  ft.  above  the  ground  with  a  velocity  of  100  ft./sec. 

2.  Construct  a  graph  showing  the  trajectory  of  a  body  projected  at  an 
angle  of  50°  with  the  ground  with  a  velocity  of  400  ft./sec. 

3.  A  projectile  is  discharged  with  a  horizontal  muzzle  velocity  100 
meters /sec.  at  a  height  of  500  meters  above  the  ground.     Find:  (a)  the 
time  of  descent;  (6)  the  vertical  velocity  on  striking;  (c)  the  random  or 
range. 

4.  A  bullet  discharged  horizontally  with  a  velocity  of  800  ft./sec. 
has  a  range  of  4,000  ft.     Find:  (a)  the  time  the  bullet  is  in  the  air;  (6) 
the  height  of  the  bullet  before  discharge. 

6.  A  ball  is  shot  from  a  cannon  at  an  angle  of  elevation  of  60°  with  an 
initial  velocity  of  1,200  meters /sec.     Find:  (a)  the  time  of  ascent;  (6) 
the  time  in  the  air;  (c)  the  final  vertical  velocity;  (d)  greatest  vertical 
height  reached;  (e)  the  range. 

6.  A  baseball  is  batted  from  the  home  plate  to  the  center  fielder,  a 
distance  of  200  ft.,  with  an  average  horizontal  velocity  of  60  ft./sec. 
Assuming  that  the  ball  starts  4  ft.  above  the  ground  and  is  caught  4  ft. 
above  the  ground,  what  is  the  greatest  vertical  height  reached  by  the 
ball? 


208  MECHANICS 

SECTION  II.     CENTRIFUGAL  FORCE 

215.  Nature  of  Centrifugal  Force. — The  particles  composing 
a  rotating  body  have  a  tendency  to  "fly  off"  in  a  straight 
line  at  a  tangent  to  the  circle  described  by  the  individual 
particle.     Suppose,  for  example,  that  we  take  particle  P  on 

the  circumference  of  a  rotating  fly 
wheel,  as  shown  in  Fig.  223.  Since 
every  body  tends  to  continue  its 
motion  in  a  straight  line  (Newton's 
first  law  of  motion),  it  is  evident  that 
P  has  a  tendency  to  leave  the  rim  of 
the  wheel  and  travel  in  a  line  shown 
by  the  arrow  A.  This  tendency, 
which  is  really  a  reaction  due  to 

a  LGdenc3y7oP"fly  off" at"  inertia'  »vw  rise  to  a  pull  away  from 

tangent"  due  to  centrifugal  the  center  of  the  circle  and  must  be 

overcome  by  an  equal  and  opposite 

force  toward  the  center.  The  force  with  which  the  particle 
tends  to  leave  the  center  is  known  as  centrifugal  force.  The 
equal  and  opposite  balancing  force  is  known  as  centripetal  force. 

216.  Effects  and  Uses  of  Centrifugal  Force.- — Centrifugal 
force  occasions  many  interesting  results,  some  of  which  are 
.made  use  of  mechanically.     Centrifugal  force  no  doubt  caused 
the  earth,  when  it  was  in  a  plastic  state,  to  become  bulged  at 
the    equator.     On    curves,    railway   tracks    are    constructed 
higher  on  the  outside  than  on  the  inside,  so  as  to  act  against 
the  overturning  tendency  of  centrifugal  force.     The  curves  at 
the  Sheepshead  Bay  automobile  track  were  built  at  a  slant  to 
prevent  skidding,  as  racing  cars  travel  at  a  speed  in  excess  of 
100  m.p.h. 

Laundries  make  use  of  centrifugal  force  in  drying  clothes. 
The  wash  is  placed  in  a  large  perforated  cylinder.  As  the 
cylinder  rotates,  the  water  tends  to  fly  out.  The  cream  sepa- 
rator operates  on  the  principle  of  centrifugal  force.  The 
milk  is  rotated  at  a  high  rate  of  speed.  The  heavier  portion 
gathers  at  the  edge  of  the  container,  leaving  the  lighter  cream 
at  the  center  where  it  is  drawn  off. 


CENTRIFUGAL  FORCE  209 

Figure  224  shows  a  model  of  a  governor  used  on  low-speed 
steam  engines  to  maintain  a  constant  fly  wheel  speed.  Due 
to  centrifugal  force,  the  balls  tend  to  move  out  as  the  speed 
of  rotation  increases.  The  movement  of  the  balls  operates  a 
lever  which  in  turn  operates  the  steam  valve,  cutting  down  the 
amount  of  steam  admitted  to  the  cylinder.  If  the  speed 
becomes  too  low,  the  balls  return  toward  the  center  of  rota- 
tion, admitting  a  greater  quantity  of  steam. 

Medium-  and   high-speed   engines   make   use   of   a   shaft 


0 


FIG.  224. — Model  of  a  steam  governor. 

governor  carried  by  the  flywheel.  It  operates  on  the  eccentric 
in  such  a  way  that  the  steam  cut-off  is  retarded  or  accelerated 
according  to  conditions.  If  the  engine  tends  to  speed  up,  the 
governor,  acting  centrifugally,  cuts  off  the  steam  sooner; 
if  the  engine  tends  to  slow  down,  the  action  is  reversed. 
Early  cut-off  decreases  the  h.p.  and  a  later  cut-off  increases 
the  h.p. 

Many  automobile  trucks  have  a  centrifugal  arrangement  to 
prevent  excessive  speed.  A  governor  is  used  which  limits 
the  supply  of  air  and  gas  delivered  to  the  cylinders.  Certain 
automobiles  have  also  a  centrifugal  device  for  regulating  the 
advance  and  retard  of  the  spark. 

14 


210  MECHANICS 

217.  Determination  of  Centrifugal  Force. — In  actual  prac- 
tice, we  are  usually  interested  in  bodies  revolving  about  some 
fixed  point.     Centrifugal  force  varies  directly  as  the  weight  of 
the  body  and  as  the  square  of  the  velocity  of  revolution; 
inversely    as    the    radius    of    revolution.     Since    centrifugal 
force  is  equal  to  centripetal  force,  one  formula  will  suffice  for 
both. 

Wv2 
Centrifugal  force  =  , 

in  which,  W  —  the  weight  of  the  body;  v  =  the  linear  velocity 
of  the  body  at  its  center  of  mass;  g  =  the  acceleration  of 
gravity;  and  r  =  the  radius  of  revolution  (the  distance  from 
the  center  of  mass  to  the  center  of  revolution) .  In  the  English 
system,  pounds  and  feet  are  used;  in  the  c.g.s.  system,  grams 
and  centimeters  are  used. 

Questions  and  Problems 

(Assume  g  =  32  ft./sec.2  or  980  cm./sec.2) 

1.  What  is  centrifugal  force?     Carefully  explain  the  cause  of  centri- 
fugal force. 

2.  What  is  centripetal  force?     How  does  it  compare  in  value  with 
centrifugal  force?     Is  centripetal  force  useful?     Explain. 

3.  Give  several  instances  showing  the  practical  application  of  centri- 
fugal force  to  mechanisms. 

4.  Give  the  formula  for  centrifugal  force. 

5.  A  body  weighing  10  Ib.  is  whirled  around  at  the  rate  of  100  r.p.m. 
If  the  radius  of  revolution  is  5  ft.,  what  centrifugal  force  is  developed? 

6.  An  automobile  weighing  3,000  Ib.  is  rounding  a  curve  at  the  rate  of 
40  m.p.h.     If  the  radius  of  curvature  is  Y±  mile,  what  friction  must  there 
be  between  the  road  and  tires  to  prevent  skidding? 

7.  A  flywheel  weighing  300  Ib.  is  rotating  1,500  times  per  minute. 
There  is  an  unbalanced  weight  of  5  Ib.  acting  3  ft.  from  the  center. 
Find  the  load  on  bearings  due  to  centrifugal  force. 

SECTION  III.     THE  PENDULUM 

218.  The  Pendulum. — The  ordinary  pendulum  consists  of  a 
bob  suspended  by  a  thread,  cord,  rod,   etc.,  and  capable  of 
oscillation  about  a  fixed  point.     If  the  weight  of  the  sup- 
porting medium  is  negligible,  the  length  of  a  pendulum  is  the 


THE  PENDULUM 


211 


distance  from  the  point  of  support  to  the  center  of  mass  of  the 
bob.  The  amplitude  is  the  linear  distance  from  the  point  of 
rest  to  the  farthest  position  of  swing.  The  period  or  time 
of  vibration  is  the  number  of  seconds  necessary  for  a  complete 
swing  over  and  back.  A  "seconds  pendulum"  is  one  which 
makes  a  half -vibration  in  one  second.  Thus,  the  period  of 
a  " seconds  pendulum"  is  2  seconds.  The  length  of  such  a 
pendulum  at  New  York  is  approximately  39  in.  or  100  cm. 
219.  Why  a  Pendulum  Vibrates. — A  vibrating  pendulum 


6 


FIG.  225. — Diagram   to  show  how  gravity  produces  the   oscillations  of  a 

pendulum. 

is  a  good  example  of  what  is  known  as  simple  harmonic  motion. 
Harmonic  motion  is  a  forward  and  back  motion,  in  which 
the  maximum  velocity  is  at  the  center,  decreasing  to  zero  at 
either  end. 

A  pendulum  set  into  vibration  will  continue  to  vibrate 
with  succeedingly  smaller  amplitudes,  until  it  comes  to  a  posi- 
tion of  rest.  The  force  tending  to  continue  the  vibration  is 
gravity  and  the  force  eventually  bringing  the  pendulum  to 
rest  is  friction. 

Let  us  see  exactly  how  gravity  acts  with  respect  to  a  pendu- 
lum. Suppose  (Fig.  225)  that  the  pendulum  bob  is  at  rest  at 


212 


MECHANICS 


B.  Gravity  (G),  acting  vertically  downward,  produces 
merely  a  tension  in  the  supporting  cord  OB.  and  there  is  no 
tendency  toward  motion.  If  now  the  bob  is  drawn  over  to  A, 
gravity  may  be  resolved  into  two  components,  AD'  and  AE'. 
AD'  produces  tension  in  OA,  while  AE',  acting  tangent  to  the 
arc  AB,  tends  to  produce  motion.  As  A  swings  to  the  right, 
AE'  decreases  in  value,  until  at  B  it  is  zero.  At  B  there  is 
sufficient  kinetic  energy  (friction  neglected)  to  carry  the  bob 
to  C.  At  C  gravity  may  again  be  resolved  into  two  com- 
ponents, CD  and  CE.  CD  produces  tension  in  OC,  while 
CE  tends  to  produce  motion.  As  C  swings  to  the  left,  CE 
decreases  in  value,  until  at  B  it  is  zero.  At  B  the  bob  has 
sufficient  kinetic  energy  (friction  neglected)  to  carry  it  to  A, 
whereupon  the  cycle  is  begun  again. 
Were  it  not  for  friction,  the  to-and-fro 
motion  would  continue  indefinitely.  In 
actual  practice,  the  arc  of  oscillation 
gradually  decreases  on  account  of  fric- 
tion, eventually  bringing  the  pendulum 
to  rest.  The  pendulum  of  a  clock  and 
the  balance  wheel  of  a  watch  receive 
energy  from  an  outside  source  (spring) 
and  continue  to  vibrate  until  the  energy 
of  the  spring  is  exhausted. 

220.  Laws  of  the  Pendulum. — The 
mass  of  a  pendulum  does  not  affect  its 
vibration  rate.  This  is  proved  by  the 
following  simple  experiment.  Referring 
to  Fig.  226,  we  have  two  pendulums  of 

FiG.226.-A^Lratus  e(lual  lenSth  but  different  weight.     They 

for  studying  laws  of  the  are  oscillated  through-  small   amplitudes 

and  the  vibrations  of  each  are  counted  for 

one  minute.     It  will  be  found  that  each  pendulum  has  made 

the  same  number  of  vibrations. 

For  small  amplitudes,  the  vibration  rate  is  independent  of 
the  arc  of  swing.  This  may  be  proved  by  oscillating  one  of  the 
pendulums  in  Fig.  226  through  different  short  arcs.  The 


THE  PENDULUM 


213 


number  of  vibrations  per  minute  will  be  the  same  in  each 
case. 

We  know  from  experience,  however,  that  the  vibration  rate 
of  a  pendulum  depends  upon  the  length 
of  the  pendulum:  the  shorter  the  length 
the  greater  number  of  vibrations  in  a 
given  time.  Suppose  (Fig.  227)  we  have 
three  pendulums,  A,  B  and  C,  of  like 
weight  and  material  and  with  lengths  of 
81, 64  and  49  cm.  respectively.  The  square 
roots  of  the  lengths  are  then  in  the  ratio 
of  9  :  8  :  7.  Each  pendulum  in  turn  is  set 
swinging  through  a  small  amplitude  and 
the  number  of  vibrations  per  minute 
counted.  The  periods  of  vibration  are 
then  determined.  It  will  be  found  that 
the  periods  are  in  the  ratio  of  9:8:7, 
proving  that  the  period  of  vibration  is 
directly  proportional  to  the  square  root  of 
the  length. 

Since  the  force  of  gravity  varies  from  "^"f  the  pendulum8 
place  to  place,  it  will  be  evident  from  Fig. 
225  that  an  increase  in  the  value  of   "g'   will  result  in  a 
decrease   in   the    period    of    vibration.     The    period   varies 
inversely  as  the  square  root  of  the  acceleration  of  gravity. 


FIG.    227. — Appa- 


LAWS  OF  THE  PENDULUM  SUMMARIZED 

1 .  The  period  of  vibration  is  independent  of  the  mass, 

2.  For  small  amplitudes,  the  period  of  vibration  is  independent 
of  the  amplitude, 

3.  The   period  of  vibration  varies  directly  as  the  square  root 
of  the  length, 

4.  The  period  of  vibration  varies  inversely  as  the  square  root 
of  the  acceleration  of  gravity. 

The  laws  in  3  and  4  above  may  be  incorporated  in  the 
following  equation : 


214  MECHANICS 

in  which  T  is  the  period  or  time  of  a  complete  vibration  in 
seconds;  I  the  length  of  the  pendulum;  and  g  the  acceleration 
due  to  gravity  at  the  particular  locality.  Thus,  knowing  the 
length  and  period  of  a  pendulum,  the  value  of  g  may  be 
readily  determined. 

Questions  and  Problems 

(Assume  g  =  32  ft. /sec.2  or  980  cm. /sec.2) 

1.  What  is  a  pendulum? 

2.  Define  the  following  terms  with  respect  to  a  pendulum:  length? 
amplitude,  period  of  vibration,  seconds  pendulum. 

3.  Show  that  a  pendulum  illustrates  simple  harmonic  motion. 

4.  What  is  the  length  of  a  seconds  pendulum  at  New  York? 
6.  State  the  laws  of  the  pendulum. 

6.  Incorporate  the  laws  of  the  pendulum  in  a  single  equation. 

7.  Two  pendulums  are  respectively  4  and  9  in.  in  length.     Compare 
their  periods  of  vibration. 

8.  The  period  of  a  pendulum  is  2  sec.     What  will  the  period  be  if 
the  length  is  doubled? 

9.  A  pendulum  makes  10  v.p.m.     How  many  v.p,m.  will  it  make  if 
the  length  is  made  one  half  as  great? 

10.  Find  the  length  of  a  second  pendulum  if  g  =  980  cm.;  990  cm. 

11.  A  suspended  plumb  line  makes   10.  complete  vibrations  in  one 
minute.     How  long  is  it?     Give  answer  in  ft. 

12.  A  plumb  line  40  ft.  long  is  dropped  from  a  chimney.     What  is 
its  period  of  vibration?     How  many  v.p.m.  does  it  make? 

13.  A  clock  pendulum  is  12  in.  long.     How  many  seconds  will  the 
clock  gain  in  24  hr.,  if  the  length  is  shortened  to  11  in.? 


APPENDIX 

USEFUL  INFORMATION 

Pi,  TT  =  3.1416. 

Circumference  of  a  circle  =  u-  X  d. 
Area  of  a  circle  =  irr2  or  d?  X  .7854. 
Area  of  a  sphere  =  4-jrr2. 

4-n-r3 
Volume  of  a  sphere  =  ~TT~' 

Lateral  surface  of  a  cylinder  =  circumference  of  base  X  altitude. 
Volume  of  a  cylinder  =  area  of  base  X  altitude. 

ENGLISH  AND  METRIC  EQUIVALENTS 

1  inch  =  2 . 54  cm.  1  meter  =  39 . 37  in. 

1  mile  =  1.61  Km.  1  kilometer  =  .62  mi. 

1  pound  =  453.6  g.  1  kilogram  =2.20  Ib. 

1  liquid  quart  =  .  946  1.  1  liter  =  1 . 057  liquid  qt. 

DENSITY  OF  COMMON  SUBSTANCES 

1  cu.  ft.  of  water  at  4°C.  weighs  62.4  Ib. 

1  cu.  in.  of  water  weighs  .  036  Ib. 

1  c.c.  of  water  at  4°C.  weighs  1  g. 

1  cu.  ft.  of  air  at  standard  conditions  weighs  .0817  Ib. 

1  c.c.  of  air  at  standard  conditions  weighs  .  00129  g. 

1  gallon  (231  cu.  in.)  of  water  weighs  8.32  Ib. 

1  cu.  ft.  of  brass  (cast)  weighs  527  Ib. 

1  cu.  ft.  of  copper  (cast)  weighs  553  Ib. 

1  cu.  ft.  of  ice  weighs  57.2  Ib. 

1  cu.  ft.  of  iron  (cast,  gray)  weighs  442  Ib. 

1  cu.  ft.  of  mercury  weighs  848  Ib. 

1  cu.  in.  of  mercury  weighs  .49  Ib. 

UNITS  FREQUENTLY  USED 

"g"  at  New  York  =  32.16  ft./sec.2 

"g"  at  New  York  =  980.2  cm./sec.2 

Length  of  a  seconds  pendulum  at  New  York  =99.3  cm, 

Idyne  =  .  00102  g. 

215 


216  MECHANICS 

1  gram  =  980  dynes. 

1  pound  =  445,000  dynes. 

1  erg  =  1  dyne  centimeter. 

1  joule  =  10,000,000  ergs. 

1  horsepower  =  33,000  ft.  Ib./min. 

1  horsepower  =  550  ft.  Ib./sec. 

1  horsepower  =  746  watts. 

1  kilowatt  =  1,000  watts. 

1  horsepower  =  746/1,000  kilowatt  (use  3/4). 

1  kilowatt  =  1,000/746  horsepower  (use  4/3). 

SPECIFIC  GRAVITY  OF  COMMON  SUBSTANCES 

SOLIDS  FLUIDS 

Aluminum  (hard  drawn) ....  2.7  Air 00129 

Brass  (cast) 8.4  Alcohol 80 

Copper  (cast) 8.9  Chloroform 1.5 

Gold  (cast) 19.3  Ether 736 

Ice 918  Gasoline 66-. 69 

Iron  (gray,  cast) 7.1  Kerosene 79 

Iron  (wrought) 7.8  Linseed  oil  (boiled) .        .  94 

Nickel 8.6  Mercury 13.6 

Platinum 21.4  Neat's  foot  oil 91 

Silver  (cast) 10 . 5  Turpentine 87 

Tin  (cast) 7.3  Water  (distilled).. .  .     1.00 

Zinc  (cast) 7. 1  Water  (sea) 1 .03 

TENSILE  STRENGTH 

Cast  iron 18,000  lb./in.2 

Wrought  iron 50,000  lb./in.2 

Mild  steel 70,000  lb./in.2 

COMPRESSIVE  STRENGTH 

Cast  iron 90,000  lb./in.2 

Wrought  iron 38,000 lb./in.2 

Mild  steel 80,000  lb./in.2 

SHEARING  STRENGTH 

Cast  iron 25,000  lb./in.2 

Wrought  iron 40,000  lb./in.2 

Mild  steel 50,000  lb./in.2 


APPENDIX 


217 


MODULUS  OF  ELASTICITY  FOR  TENSION  AND  COMPRESSION 

Cast  iron 15,000,000  lb./in.2 

Wrought  iron 25,000,000  lb./in.2 

Mild  steel 30,000,000  lb./in.2 

MODULUS  OF  ELASTICITY  FOR  SHEAR 

Cast  iron 6,000,000  lb./in.2 

Wrought  iron 10,000,000  lb./in.2 

Mild  steel 12,000,000  lb./in.2 

DECIMAL  EQUIVALENT  PARTS  OF  AN  INCH 

K4 0156  3%4 5156 

H2 0313  i%2 5313 

%4 0469  3%4 5469 

He 0625  %6 5625 

%4 0781  3%4 5781 

%2 0938  i%a 5938 

%4 1094  3%4 6094 

H  125  %  625 

%4 1406  4K4 6406 

%2 1563  2i^2 6563 

1719  4%4 6719 

1875    1^6 6875 

2031    4%4 7031 

2188    2^2 7188 

2344  4%4 7344 

25      M 75 

2656  4%4 7656 

2813  2^2 7813 

2969  s^4 7969 

3125  ijf6 8125 

3281  s%4 8281 

3438  2%2 8438 

3594  5%4 8594 

375     y8  875 

-3906  5%4 8906 

4063  2%2 9063 

%4 4219  5%4 9219 

Ke 4375  % 9375 

4531  e^4 9531 

4688  3J^2 9688 

4844  6^4 9844 

.51..  .1 


218 


MECHANICS 
TRIGONOMETRIC  FUNCTIONS 


A 

Sin 

Cos 

Tan 

A 

Sin 

Cos 

Tan 

0 

0.000 

1.000 

0.000 

1 

0.017 

0.999 

0.017 

46 

0.719 

0.695 

1.04 

2 

0.035 

0.999 

0.035 

47 

0.731 

0.682 

1.07 

3 

0.052 

0.999 

0.  052 

48 

0.743 

0.669 

.11 

4 

0.070 

0.998 

0.070 

49 

0.755 

0.656 

.15 

5 

0.087 

0.996 

0.087 

50 

0.766 

0.643 

.19 

6 

0.105 

0.995 

0.105 

51 

0.777 

0.629 

.23 

7 

0.122 

0.993 

0.123 

52 

0.788 

0.616 

.28 

8 

0.139 

0.990 

0.141 

53 

0.799 

0.602 

.33 

9 

0.156 

0.988 

0.158 

54 

0.809 

0.588 

.38 

10 

0.174 

0.985 

0.176 

55 

0.819 

0.574 

.43 

11 

0.191 

0.982 

0.194 

56 

0.829 

0.559 

.48 

12 

0.208 

0.978 

0.213 

57 

0.839 

0.545 

.54 

13 

0.225 

0.974 

0.231 

58 

0.848 

0.530 

.60 

14 

0.242 

0.970 

0.249 

59 

0.857 

0.515 

.66 

15 

0.259 

0.966 

0.268 

60 

0.866 

0.500 

.73 

16 

0.276 

0.961 

0.287 

61 

0.  875 

0.485 

.80 

17 

0.292 

0.956 

0.306 

62 

0.883 

0.469 

.88 

18 

0.309 

0.951 

0.325 

63 

0.891 

0.454 

.96 

19 

0.326 

0.946 

0.344 

64 

0.898 

0.438 

2.05 

20 

0.342 

0.940 

0.364 

65 

0.906 

0.423 

2.14 

21 

0.358 

0.934 

0.384 

66 

0.914 

0.407 

2.25 

22 

0.375 

0.927 

0.404 

67 

0.921 

0.391 

2.36 

23 

0.391 

0.921 

0.424 

68 

0.927 

0.375 

2.48 

24 

0.407 

0.914 

0.445 

69 

0.934 

0.358 

2.61 

25 

0.423 

0.906 

0.466 

70 

0.940 

0.342 

2.75 

26 

0.438 

0.898 

0.  488 

71 

0.946 

0.326 

2.90 

27 

0.454 

0.891 

0.510 

72 

0.951 

0.309 

3.08 

28 

0.  469 

0.883 

0.532 

73 

0.956 

0.292 

3.27 

29 

0.485 

0.875 

0.554 

74 

0.961 

0.276 

3.49 

30 

0.500 

0.866 

0.577 

75 

, 

0.966 

0.259 

3.73 

31 

0.515 

0.857 

0.601 

76 

0.970 

0.242 

4.01^ 

32 

0.530 

0.848 

0.625 

77 

0.974 

0.225 

4.33 

33 

0.545 

0.839 

0.649 

78 

0.978 

0.208 

4.70 

34 

0.559 

0.829 

0.675 

79 

0.982 

0.191 

5.14 

35 

0.574 

0.819 

0.700 

80 

0.985 

0.174 

5.67 

36 

0.588 

0.809 

0.727 

81 

0.988 

0.156 

6.31 

37 

0.602 

0.799 

0.754 

82 

0.990 

0.139 

7.12 

38 

0.616 

0.788 

0.781 

83 

0.993 

0.122 

8.14 

39 

0.629 

0.777 

0.810 

84 

0.995 

0.105 

9.51 

40 

0.643 

0.766 

0.839 

85 

0.996 

0.087 

11.43 

41 

0.656 

0.755 

0.869 

86 

0.998 

0.070 

14.30 

42 

0.669 

0.743 

0.900 

87 

0.999 

0.052 

19.08 

43 

0.682 

0.731 

0.933 

88 

0.999 

0.035 

28.64 

44 

0.695 

0.719 

0.966 

89 

0.999 

0.017 

57.28 

45 

0.707 

0.707 

1.000 

90 

1.000 

0.000 

Infinity 

INDEX 


Absolute  motion,  35 
Acceleration,  36 

of  gravity,  25,  37,  203 
Acme  thread,  147 
Adhesion,  5 
Aeroplane,  190 
Air,  180 

brake,  194 

chamber,  196 

pump,  193 
American  system  of  rope 

mission,  155 
Amplitude,  211 
Aneroid  barometer,  184 
Annealing,  6 

Annular  ball  bearing,  108 
Anti-friction  devices,  106 
Apex,  67 
Archimedes,  177 

law  of,  177 
Areas,  215 
Arm,  couple,  60 

moment,  33 
Ascensional  force,  191 
Atmosphere,  180 
Atmospheric  pressure,  180 
Atom,  2 

Automobile,  135 
Axis  of  rotation,  32,  37 

B 

Babbitt  metal,  106 
Ball  bearing,  106,  107 
Balloon,  191 


Baltimore  truss,  68 
Barograph,  185 
Barometer,  mercurial,  183 

aneroid,  184 
Bearing,  106 

Bearings,  lubrication  of,  110 
Belts,  150 

Bevel  gear,  155,  156,  157 
Bicycle,  130 
Block,  124 

Block  and  tackle,  117 
Block,  chain,  148 
trans-      Bole,  38 
Boom,  73 
Bottom  slack,  150 
Boyle,  Robert,  188 
Boyle's  law,  188 
Brace,  67 
Brake,  prony,  93 

horsepower,  93 
Bridge,  67 

truss,  67 

British  thermal  unit,  100 
Brittleness,  6 
Buoyant  force,  176 
Butt  joint,  84 


Caliper,  12,  13 

micrometer,  14,  15 
vernier,  16 
Cam,  145 
follower,  141 
shaft,  141 

Capacity,  metric  table  of,  10 
units  of,  9 


219 


220 


INDEX 


Center  of  gravity,  25 

pressure,  170 
Centigram,  8 
Centiliter,  9 
Centimeter,  8 
Centrifugal  force,  208 

pump,  197 

Centripetal  force,  208 
C.g.s.  system,  7 
Chain  and  sprocket,  147 
Chain  block,  124 
Chemical  change,  3 
Chord,  67 
Circle,  area,  215 

circumference,  215 
Clutch,  144 

Coefficient  of  friction,  104 
Cohesion,  4 
Component,  44 

horizontal,  49 

vertical,  49 

Components,  rectangular,  49 
Composition  of  forces,  44,  65 

velocities,  46 
Compound  drive,  151 

truss,  66 

Compressibility,  4 
Compression,  32,  64,  66 

member,  66 

Compressive  strength,  table,  216 
Concurrent  forces,  52 
Connecting  rod,  106 
Conservation  of  energy,  100 
Continuous  system  of  rope  trans- 
mission, 155 
Cosecant,  20 
Cosine,  20,  218 
Cotangent,  20 
Countershaft,  136,  141 
Couple,  60 

arm,  60 
Coupling,  142 
Crane,  64 

hoisting,  73 


Crankshaft,  140 
Cream  separator,  208 
Crow  bar,  114 
Cup-and-cone  bearing,  107 
Curved-tooth  gear,  157 
Curvilinear  motion,  36 
Cutting  tools,  122 
Cylinder,  lateral  area,  215 
volume,  215 

D 

Dam,  171 

Day,  mean  solar,  9 

Dead  load,  66 

Decigram,  8 

Deciliter,  9 

Decimal  equivalents,  217 

Decimeter,  8 

Deck  truss,  66 

Dekameter,  9 

Density,  162 

table  of,  215 

of  water,  9 
Diagonal,  67 
Diagram,  engine,  90 

indicator,  90 
Differential  block,  124 
Double  shear,  85 
Ductility,  6 
Dyne,  87 

E 

Efficiency,  113,  123,  126,  128,  131, 
134 

mechanical,  96 
Elastic  fatigue,  83 

limit,  79 
Elasticity,  4,  77 

modulus  of,  81,  217 
Electron,  2 
Energy,  2,  98 

conservation  of,  100 

kinetic,  99 


INDEX 


221 


Energy,  potential,  98 

transformation  of,  100 
Engine  diagram,  90 
English    and    metric    equivalents, 

215 
system  of  measurement,  7 

rope  transmission,  154 
Equilibrant,  46 
Equilibrium,  3,  27,  52,  57,  62 
Equivalents,  metric  and  English, 

10,  215 

of  heat,  mechanical,  100 
Erg,  87 


Factor  of  safety,  82 
Fafnir  ball  bearing,  108 
Failure  of  rivets,  85 
Falling  bodies,  202 
Fatigue,  elastic,  83 
Female  thread,  147 
Fink  truss,  67 
First  class  lever,  114 
Fishing  rod,  115 
Fixed  coupling,  142 

pulley,  117 
Flexible  coupling,  143 

shaft,  142 
Foot-pound,  88 
Force,  30,  35 

centripetal,  208 

centrifugal,  208 

graphical  representation  of,  31 

measure  of,  31 

moment  of,  32 

of  gravity,  24 

pump,  195 
Forces,  composition  of,  44,  65 

concurrent,  52 

non-concurrent,  62 

parallelogram  of,  45 

triangle  of,  53 
F.p.s.  system,  7 


Friction,  103 

clutch,  144 

coefficient  of,  104 

laws  of,  105 

sliding,  104 

wheels,  155 

Fulcrum,  32,  57,  114,  115 
Functions,  trigonometric,  20,  218 
Fundamental  units,  7 


G 


"g,"  215 

Galileo,  Galilei,  182 
Gas,  161,  180 
Gauge,  pressure,  198 

vacuum,  198 
Gear,  156 

transmission,  135 

reverse,  136 

pump,  197 

Geometrical  formulas,  215 
Governor,  209 
Gram,  8 

Graph,  126,  129,  132,  135 
Graphical  representation  of  forces, 

31 
Gravitation,  23 

law,  23 
Gravity,  24 

acceleration  of,  25,  37,  203 
Ground  reaction,  63 


II 


Hardness,  5 

Heat,  mechanical  equivalent,  100 

Hectometer,  9 

Helical  gear,  156,  157,  159 

Herringbone  gear,  159 

High-duty  bearing,  106 

Highway  bridge,  67 

Hoisting  crane,  73 

Hollow  shaft,  140 


222 


INDEX 


Hooke,  Robert,  77 
Hooke's  law,  77,  82 
Horizontal,  24 

component,  49 
Horsepower,  92,  152 

indicated,  95 
Howe  truss,  67 
Hyatt  bearing,  109 
Hydraulic  press,  166 
Hydrometer,  178 


Idler,  150,  151 

reverse,  137 

Inclined  plane,  112,  120 
Indestructibility,  4 
Indicated  horsepower,  95 
Indicator,  96 

diagram,  90 
Inertia,  4 

law  of,  39 
Input,  112 
Inside  caliper,  13 
Instantaneous  velocity,  37 


Jack  screw,  121,  132 
Joint,  84 
Joule,  88 

James  Prescott,  101 

K 

Kelvin,  Lord,  2 
Kilogram,  8 
Kilogram-meter,  88 
Kilowatt,  92,  93 
Kinetic  energy,  99 

theory,  3 
Kinetics,  1 


Ladder,  62 


Lap  joint,  84 

Lathe,  lead  screw,  147 

Law,  Archimedes',  177 

Boyle's,  188 

gravitation,  23 

Hooke's,  82 

inertia,  39 

machines,  113 

moments,  57 

parallelogram,  44 

Pascal's,  165 
Laws,  friction,  105 

liquid  pressure,  164 

motion,  38 

pendulum,  213 
Lead,  121,  147 
Length,  measures  of,  9 

standard,  English,  7 

metric,  8 

Lever,  112,  114,  115 
Lift,  191 

pump,  195 
Line  shaft,  140 
Linear  table,  metric,  9 
Link,  146 

liquid,  161 
Liquid  pressure,  computation,  170 

laws,  164 
Liter,  9 
Live  load,  66 
Load,  dead,  66 
Low-duty  bearing,  106 
Lubrication,  110 

M 

Machines,  112 

laws  of,  113 

Magdeburg  hemispheres,  186 
Main  shaft,  140 
Male  thread,  146 
Malleability,  6 

Manila  transmission  rope,  154 
Manometer,  199 


INDEX 


223 


Mass,  5 

standard  of,  English,  7 

metric,  8 
Matter,  2 
Mean  solar  day,  9 
Measurement  of  a  force,  31 
Measures,  metric,  tables  of,  9,  10 
Measuring  instruments,  11 
Mechanical  advantage,  113 

efficiency,  96 

equivalent  of  heat,  100 
Mechanics,  1 

of  liquids,  163 
Medium-duty  bearing,  106 
Member,  compression,  66 

tension,  66 

Mercurial  barometer,  183 
Meter,  8 
Metric    and   English   equivalents, 

215 
tables,  9,  10 

system,  7 

Micrometer  caliper,  14,  15 
Milligram,  8 
Milliliter,  9 
Millimeter,  8 
Mitre  gear,  157 

Modulus,   of   elasticity,    Young's, 
81,  217 

of  rigidity,  84 
Molecule,  2 
Moment  of  a  force,  32 

arm,  33 

Moments,  law  of,  57 
Momentum,  38 
Motion,  35 

laws  of,  Newton's,  38 
Multiple    system    of    rope    trans- 
mission, 154 
Myriameter,  9 


N 


Neutral  equilibrium,  27 


Newton,  Sir  Isaac,  23 
Newton's  laws  of  motion,  38 
Non-concurrent  forces,  62 


Output,  113 

Outside  caliper,  12,  13 

Overshot  wheel,  174 


Panel,  67 
Parallel  forces,  57 
Parallelogram  law,  44 

of  forces,  45 
Pascal's  law,  165 
Pelton  wheel,  175 
Pendulum,  210 

laws  of,  213 
Penstock,  175 
Period,  211 
Physical  change,  3 
Physics,  1 
Pinion,  157 
Pitch,  121,  147 
Planetary  system,  130 
Plumb  line,  24 
Porosity,  3 
Positive  clutch,  144 
Potential  energy,  98 
Power,  92 
Pratt  truss,  67 
Pressure,  162 

atmospheric,  180 

gauge,  198 

liquid,  163 
Projectiles,  204,  205 
Prony  brake,  93 
Properties  of  matter,  3 
Protractor,  12 
Pulley,  112,  117 

whip-on-whip,  139 
Pulleys  and  ropes,  153 


224 


INDEX 


Pump,  air,  193 
force,  195 
lift,  195 

reciprocating,  195 
rotary,  197 


Quadrangular  truss,  68 
R 

Radial  bearing,  106 
Radian,  41 
Railroad  bridge,  67 
Range,  206 
Reaction,  31 

ground,  63 

Reciprocating  pump,  195 
Rectangular  components,  49 
Rectilinear  motion,  36 
Representation  of  forces,  graphic, 

31 
Resolution  of  forces,  48 

velocities,  48 
Resultant,  44,  46 

parallel  forces,  59 
Retaining  walls,  171 
Reverse  gear,  136 

idler,  137 

Rigidity,  modulus  of,  84 
Rivets,  failure  of,  85 
Roller  bearing,  106,  109 

chain,  148 
Roof  truss,  67,  68 
Rotary  motion,  36 

pump,  197 
Rotation,  36 

axis  of,  32 


S 


Safety  factor,  82 
Screw,  112,  121 


Screw,  threads,  146 
Screw-geared  block,  124,  127 
Secant,  20 

Second  class  lever,  115 
Secondary  shaft,  142 
Shaft,  140 

counter,  136 

spline,  136 
Shear,  83 

double,  85 

single,  85 
Shear  legs,  74 
Shearing  strain,  84 

strength,  table,  216 

stress,  84 
Silent  chain,  149 
Simple  harmonic  motion,  211 

machines,  112 
Sine,  20,  218 
Siphon,  193 
Slide  rule,  14 
Sliding  friction,  104 

gear,  135 

Society  Automotive  Engineers,  95 
Solid,  161 

Specific  gravity,  162,  178,  216 
Speed,  37 

indicator,  13,  14 
Sphere,  area,  215 

volume,  215 
Spiral  gear,  156,  159 
Spline  shaft,  136 
Sprocket,  147 
Spur  gear,  155,  156,  167 
Spur-geared  block,  124,  129,  130 
Square  thread,  147 
Stability,  28 
Stable  equilibrium,  27 
Standard  of  length,  English,  7 

metric,  8 
mass,  English,  7 

metric,  8 
time,  7 

conditions,  186 


INDEX 


225 


Statics,  1 

Stay,  67 

Steam  engine  diagram,  90 

Steel  rule,  11 

Stick  and  tie,  70 

Strain,  81 

shearing,  84 
Stress,  80 

shearing,  84 
Strut,  67 

Surface,  metric  table,  10 
System,  c.g.s.,  7 

f.p.s.,  7 

metric,  7 


Trigonometry,  20 
Trusses,  66 
Turbine  wheel,  175 
Turnbuckle,  138 

U 

Undershot  wheel,  175 
Units,  7 

of  power,  92,  93 

of  work,  87 
Universal  coupling,  143 

gravitation,  23 
Unstable  equilibrium,  27 


Tables,  215 

metric,  9,  10 
Tail  race,  176 
Tandem  drive,  151 
Tangent,  20,  218 
Tempering,  6 
Tenacity,  5 

Tensile  strength,  5,  82,  216 
Tension,  32,  64,  66 
Theory,  kinetic,  3 
Third  class  lever,  115 
Thomson,  Sir  William,  2 
Thrust  bearing,  106 
Tie,  67 

Time,  standard  of,  7 
Timken  bearing,  109,  110 
Tools,  cutting,  122 
Toothed  gear,  156 
Top  slack,  150 
Torricelli,  182 
Trajectory,  206 
Transformation  of  energy,  100 
Translatory  motion,  36 
Transmission,  140 

gear,  135 

Triangle  of  forces,  53 
Trigonometric  functions,  20,  218 


Vacuum,  182,  202 

gauge,  198 
Valves,  141 
Vector,  41 

Velocities,  composition  of,  46 
Velocity,  37 

instantaneous,  37 

ratio,  113,  125,  128,  131,  134 
Vernier  caliper,  16 
Vertical,  24,  67 

component,  49 
Vise,  138 
Volume,  metric  table,  10 

of  a  cylinder,  215 

of  a  sphere,  215 
Von  Guericke,  Otto,  186 
"V"  thread,  147 

W 

Wall  crane,  64 
Water,  density  of,  9 

wheel,  174 
Watt,  93 

James,  92 
Web  member,  67 
Wedge,  112,  122 


226  INDEX 

Worm,  127 

W**£!*M      f   in  gear,  156,  168 

metric  table  of,  10  h    i    127 

Wheel  and  axle,  112,  119 

Wheelbarrow,  115  Y 

"Whip-on-whip"  pulley,  139 

Wire  transmission  rope,  154  Yard,  8 

Work,  87  Yield  point,  80 

diagram,  89  Young's  modulus  of  elasticity,  81, 

units,  87  217 


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STAMPED  BELOW 

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NOV  2  9  1968 
NOV  24  REC'Q 

UCD  LIBRARY 
DUE  J  UN  14  1969 

JKM5    flEC'O 

J  UL> 

MAY  3  i  1971 
MAY  2  5  REC'D 


LIBRARY,  UNIVERSITY  OF  CALIFORNIA,  DAVIS 

BookSlip-JJ5ni-6,'66(G3S55sJ)458 


1647 

Smith 


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Indus  trl 


al  physics; 


Mechanics. 


£*  ' 


UNIVER3JTY  OF  CALIFORNIA  LIBRARY 


